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Course: 8th grade   >   Unit 4

Systems of equations with graphing.

  • Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
  • Systems of equations with graphing: 5x+3y=7 & 3x-2y=8
  • Systems of equations with graphing: chores
  • Systems of equations with graphing: exact & approximate solutions

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How to Solve Systems of Equations by Graphing

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  • Solve Systems of Equations by Graphing

Overview of Solve Systems of Equations by Graphing:

How does graphing help solve a system of equations, solving a system of equations by graphing, graphing equations using slope and y-intercept and solving: case 1, graphing equations using slope and y-intercept and solving: case 2, graphing equations using x- and y-intercepts and solving, it’s your turn now, the gist of what we learned so far, graphing tool.

  • Solving Systems of Equations by Graphing - Quiz

When we graph a linear equation on a coordinate plane, we get a straight line. You might wonder how this would help solve a system of linear equations. Now, look at this system.

When the equations are graphed on a grid, we’ll have:

Graph_1

When we consider each line individually, every point on it is the solution of respective equation. But, we need to look for a solution that satisfies both equations simultaneously.

What do you think would be the solution of the system?

You’re right! The point of intersection of the two lines is the solution of the system. Here, the system has a unique solution and that would be (−1, −1) .

What if the lines don’t intersect?

There are two situations, we must consider.

✯ When the lines are parallel, the system has no solution .

✯ When both equations represent the same line, the system has infinitely many solutions .

In this lesson, we’ll deal with graphing and solving systems of equations that have a unique solution .

Let’s look at the step-by-step process of solving a linear system by graphing.

Step 1: Analyze what form each equation of the system is in.

Step 2: Graph the equations using the slope and y-intercept or using the x- and y-intercepts.

  Case 1: If the equations are in the slope-intercept form, identify the slope and y-intercept and graph them.

  Case 2: If one of the equations is in slope-intercept form, rewrite the other one too in that form and graph them.

  Case 3: If both the equations are in other forms, find the x- and y-intercepts and graph them.

Step 3: The ordered pair of the point where the two lines intersect is the required solution.

Let’s explore solving systems of equations in each of the cases.

Here’s a linear system:

The given equations are already in the form y = mx + b, where m is the slope and b is the y-intercept. So, let’s move on to graphing them.

To graph the first equation, we need to use the slope and y-intercept.

The y-intercept in this case is 2. So, let’s plot the point (0, 2) on the coordinate plane.

Graph_2

Now, the slope is −1. So, let’s move down 1 unit and right 1 unit and plot another point.

Graph_3

When we connect these two points drawing a line, we get the graph of the first equation.

Graph_4

Let’s graph the second equation in the similar manner, too.

The y-intercept is 6. So, we need to plot (0, 6) on the grid.

The slope is 1. So, we must move up 1 unit and right 1 unit and plot the other point.

Drawing a line connecting the points, we have:

Graph_5

As you can see, both the lines intersect at a point. Let's label it on our graph.

Graph_6

Now, the ordered pair that corresponds to the point of intersection of the equations is the required solution.

Thus, the solution of the system of equations is (−2, 4) .

Solve y = −x + 7 and y = 2x 3 − 3 graphically.

Both equations are in the slope-intercept form.

Equation 1: m = −1; b = 7

Equation 2: m = 2 3 ; b = −3

Representing the equations graphically we have:

Graph_7

The point of intersection of the two lines is (6, 1).

The solution is (x, y) = (6, 1) .

Check Your Solution! You can cross-check if your solution is right by substituting the x and y values in the equations. Plugging (6, 1) in equation 1, you get: 1 = −6 + 7   1 = 1 ✔ Plugging (6, 1) in equation 2, you get: 1 = 2 (6) 3 − 3 1 = 12 3 − 3 1 = 4 − 3   1 = 1 ✔

Look at this system:

y = −x 2 + 5 2

The former is in the standard form and the latter is in the slope-intercept form. Let’s rearrange and rewrite equation 1 in the form y = mx + c.

Subtracting x from both sides:

 3y = −x

Dividing both sides by 3:

 y = −x 3

So, equation 1 has a slope of −1 3 and y-intercept 0.

Equation 2 has a slope of −1 2 and y-intercept of 5 2 .

Representing the system graphically, we have:

Graph_8

As you can see, the lines cut each other at (15, −5).

Thus, the solution of this system is (x, y) = (15, −5) .

Note: You can also graph the second equation using the x- and y-intercepts rather than rewriting them in slope-intercept form.

If both the equations are rendered in standard form as this system:

−8x + y = −4

We’ll find the x and y-intercepts and solve this system. When we plug y = 0 in the equations, we’ll get the x-intercepts; similarly, when we set x =0, we’ll get the y-intercepts.

Plugging y = 0 in equation 1, we get:

−8x + 0 = −4

−8x −8 = −4 −8   [Dividing both sides by −8]

⇨ x = 1 2

The x-intercept is ( 1 2 , 0) .

Plugging x = 0 in equation 1, we get:

−8(0) + y = −4

⇨ y = −4

The y-intercept is (0, −4) .

Plugging y = 0 in equation 2, we get:

⇨ x = −3

The x-intercept is (−3, 0) .

Plugging x = 0 in equation 2, we get:

−y −1 = −3 −1   [Dividing both sides by −1]

⇨ y = 3

The y-intercept is (0, 3) .

Let’s plot the x- and y-intercepts and graph both the equations.

Graphing equation 1, we get:

Graph_9

Graphing equation 2, we get:

Graph_10

As you can see, both the lines intersect at (1, 4).

Therefore, the solution is (x, y) = (1, 4) .

Note: You can also graph the system by rewriting the equations in slope-intercept form.

Thus, no matter what form the equations are in, all you need to do is graph them and identify the point of intersection.

Refresh

Use the interactive graph below to find the solution of this system.

y = 2x – 3

y = 3x – 2

Drag the point on the line and drop it on the spot you want to plot it.

Solve 2x + y = −1 and y = −x 3 + 4 graphically and check your solution.

Equation 1:

x-intercept = −1 2 ; y-intercept = −1

Equation 2:

Slope = −1 3 ; y-intercept = 4

Graph_11

The solution is (−3, 5) .

Hide answer

View answer

The solution of a system of equations is the point of intersection of the lines obtained when the equations are graphed.

When the lines don’t intersect and are parallel, the system has no solution.

When both the equations give the same line, the system has infinitely many solutions.

No matter which method you use to graph, the solution of a system is unique.

Sample Worksheets

Scale up your skills with our free printable Solving Systems of Linear Equations worksheets !

Use this graphing tool to solve the linear systems in the quiz section. This comes with a provision to alter the scale. Move the points and lines, graph the equations, and solve the system.

Take a Quiz on Solving Systems of Equations by Graphing Now!

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  • 4.1 Solve Systems of Linear Equations with Two Variables
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Determine whether an ordered pair is a solution of a system of equations
  • Solve a system of linear equations by graphing
  • Solve a system of equations by substitution
  • Solve a system of equations by elimination
  • Choose the most convenient method to solve a system of linear equations

Be Prepared 4.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 , y = 2 3 x − 4 , ⓐ Is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ Is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 3.2 .

Be Prepared 4.2

Find the slope and y -intercept of the line 3 x − y = 12 . 3 x − y = 12 . If you missed this problem, review Example 3.16 .

Be Prepared 4.3

Find the x- and y -intercepts of the line 2 x − 3 y = 12 . 2 x − 3 y = 12 . If you missed this problem, review Example 3.8 .

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

A linear equation in two variables, such as 2 x + y = 7 , 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ) . ( x , y ) .

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example 4.1

Determine whether the ordered pair is a solution to the system { x − y = −1 2 x − y = −5 . { x − y = −1 2 x − y = −5 .

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

Determine whether the ordered pair is a solution to the system { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system { x − 3 y = −8 − 3 x − y = 4 . { x − 3 y = −8 − 3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

  • Step 1. Graph the first equation.
  • Step 2. Graph the second equation on the same rectangular coordinate system.
  • Step 3. Determine whether the lines intersect, are parallel, or are the same line.
  • If the lines intersect, identify the point of intersection. This is the solution to the system.
  • If the lines are parallel, the system has no solution.
  • If the lines are the same, the system has an infinite number of solutions.
  • Step 5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts.

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Solve the system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Example 4.5

Solve the system by graphing: { y = 2 x − 3 − 6 x + 3 y = − 9 . { y = 2 x − 3 − 6 x + 3 y = − 9 .

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident Lines

Coincident lines have the same slope and same y- intercept.

The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.

In Example 4.5 , the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4.4 , has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1 .

Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ { y = 3 x − 1 6 x − 2 y = 12 { y = 3 x − 1 6 x − 2 y = 12 ⓑ { 2 x + y = − 3 x − 5 y = 5 { 2 x + y = − 3 x − 5 y = 5

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Try It 4.11

ⓐ { y = −2 x − 4 4 x + 2 y = 9 { y = −2 x − 4 4 x + 2 y = 9 ⓑ { 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 4.12

ⓐ { y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6 ⓑ { x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

Solve a system of equations by substitution.

  • Step 1. Solve one of the equations for either variable.
  • Step 2. Substitute the expression from Step 1 into the other equation.
  • Step 3. Solve the resulting equation.
  • Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
  • Step 5. Write the solution as an ordered pair.
  • Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .

Then rewrite the system of equations.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 4.9

How to solve a system of equations by elimination.

Solve the system by elimination: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

  • Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.
  • Step 3. Add the equations resulting from Step 2 to eliminate one variable.
  • Step 4. Solve for the remaining variable.
  • Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
  • Step 6. Write the solution as an ordered pair.
  • Step 7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ − 2 x + 4 y = 4 y = 1 2 x { − 2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: { 3 x + y = 12 x = y − 8 . { 3 x + y = 12 x = y − 8 .

Solve the system of equations by elimination and explain all your steps in words: { 5 x + 4 y = 10 2 x = 3 y + 27 . { 5 x + 4 y = 10 2 x = 3 y + 27 .

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing ⓑ by substitution ⓒ Which method do you prefer? Why?

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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  • Authors: Lynn Marecek, Andrea Honeycutt Mathis
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  • Book title: Intermediate Algebra 2e
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  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/4-1-solve-systems-of-linear-equations-with-two-variables

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4. Systems of Equations

4.1 Solve Systems of Equations by Graphing

Lynn Marecek and MaryAnne Anthony-Smith

Learning Objectives

By the end of this section it is expected that you will be able to:

  • Determine whether an ordered pair is a solution of a system of equations
  • Solve a system of linear equations by graphing
  • Determine the number of solutions of linear system
  • Solve applications of systems of equations by graphing

Determine Whether an Ordered Pair is a Solution of a System of Equations

 We learned before how to solve linear equations with one variable. Now we will work with systems of linear equations , two or more linear equations grouped together, witch is known as a system of linear equations.

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations.

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}

A linear equation in two variables, like 2 x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) that make both equations true. These are called the solutions to a system of equations .

Solutions of a System of Equations

Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ).

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Let’s consider the system below:

\left\{\begin{array}{c}3x-y=7\hfill \\ x-2y=4\hfill \end{array}

The ordered pair (2, −1) made both equations true. Therefore (2, −1) is a solution to this system.

Let’s try another ordered pair. Is the ordered pair (3, 2) a solution?

This figure begins with the sentence, “We substitute x equals 3 and y equals 2 into both equations.” The first equation reads 3 times x minus 7equals 7. Then, 3 times 3 minus 2 equals 7. Then 7 = 7 is true. The second equation reads x minus 2y equals 4. The n times 2 minus 2 times 2 = 4. Then negative 2 = 4 is false.

The ordered pair (3, 2) made one equation true, but it made the other equation false. Since it is not a solution to both equations, it is not a solution to this system.

\left\{\begin{array}{c}x-y=-1\hfill \\ 2x-y=-5\hfill \end{array}

  • a) yes b) no

Equations by Graphing

In this chapter we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown in (Figure 1) :

This figure shows three x y-coordinate planes. The first plane shows two lines which intersect at one point. Under the graph it says, “The lines intersect. Intersecting lines have one point in common. There is one solution to this system.” The second x y-coordinate plane shows two parallel lines. Under the graph it says, “The lines are parallel. Parallel lines have no points in common. There is no solution to this system.” The third x y-coordinate plane shows one line. Under the graph it says, “Both equations give the same line. Because we have just one line, there are infinitely many solutions.”

For the first example of solving a system of linear equations in this section and in the next two sections, we will solve the same system of two linear equations. But we’ll use a different method in each section. After seeing the third method, you’ll decide which method was the most convenient way to solve this system.

\left\{\begin{array}{c}2x+y=7\hfill \\ x-2y=6\hfill \end{array}.

The steps to use to solve a system of linear equations by graphing are shown below

To solve a system of linear equations by graphing.

  • Graph the first equation.
  • Graph the second equation on the same rectangular coordinate system.
  • Determine whether the lines intersect, are parallel, or are the same line.
  • If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. This is the solution to the system.
  • If the lines are parallel, the system has no solution.
  • If the lines are the same, the system has an infinite number of solutions.

\left\{\begin{array}{c}y=2x+1\hfill \\ y=4x-1\hfill \end{array}.

Usually when equations are given in standard form, the most convenient way to graph them is by using the intercepts. We’ll do this in the next example.

\left\{\begin{array}{c}x+y=2\hfill \\ x-y=4\hfill \end{array}.

We will find the x – and y -intercepts of both equations and use them to graph the lines.

\left\{\begin{array}{c}x+y=6\hfill \\ x-y=2\hfill \end{array}.

Do you remember how to graph a linear equation with just one variable? It will be either a vertical or a horizontal line.

\left\{\begin{array}{c}y=6\hfill \\ 2x+3y=12\hfill \end{array}.

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

\left\{\begin{array}{c}y=\frac{1}{2}x-3\hfill \\ x-2y=4\hfill \end{array}.

  • no solution

\left\{\begin{array}{c}y=2x-3\hfill \\ -6x+3y=-9\hfill \end{array}.

  • infinitely many solutions

If you write the second equation in Example 8 in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident. Coincident lines have the same slope and same y -intercept.

Coincident Lines

Coincident lines have the same slope and same y -intercept.

Determine the Number of Solutions of a Linear System

There will be times when we will want to know how many solutions there will be to a system of linear equations, but we might not actually have to find the solution. It will be helpful to determine this without graphing.

We have seen that two lines in the same plane must either intersect or are parallel. The systems of equations in Example 2 through Example 6 all had two intersecting lines. Each system had one solution.

A system with parallel lines, like Example 7, has no solution. What happened in Example 8? The equations have coincident lines , and so the system had infinitely many solutions.

We’ll organize these results in (Table 1) below:

This table has two columns and four rows. The first row labels each column “Graph” and “Number of solutions.” Under “Graph” are “2 intersecting lines,” “Parallel lines,” and “Same line.” Under “Number of solutions” are “1,” “None,” and “Infinitely many.”

Parallel lines have the same slope but different y -intercepts. So, if we write both equations in a system of linear equations in slope–intercept form, we can see how many solutions there will be without graphing! Look at the system we solved in Example 7.

\begin{array}{cccc}& & & \hfill \phantom{\rule{0.1em}{0ex}}\left\{\phantom{\rule{0.1em}{0ex}}\begin{array}{ccc}\hfill y& =\hfill & \frac{1}{2}x-3\hfill \\ \hfill x-2y& =\hfill & 4\hfill \end{array}\hfill \\ \text{The first line is in slope-intercept form.}\hfill & & & \text{If we solve the second equation for}\phantom{\rule{0.2em}{0ex}}y,\phantom{\rule{0.2em}{0ex}}\text{we get}\hfill \\ \hfill y=\frac{1}{2}x-3\hfill & & & \hfill \phantom{\rule{1em}{0ex}}\begin{array}{ccc}\hfill x-2y& =\hfill & 4\hfill \\ \hfill -2y& =\hfill & \text{−}x+4\hfill \\ \hfill y& =\hfill & \frac{1}{2}x-2\hfill \end{array}\hfill \\ \hfill m=\frac{1}{2},b=-3\hfill & & & \hfill m=\frac{1}{2},b=-2\hfill \end{array}

The two lines have the same slope but different y -intercepts. They are parallel lines.

(Table 2) shows how to determine the number of solutions of a linear system by looking at the slopes and intercepts.

This table is entitled “Number of Solutions of a Linear System of Equations.” There are four columns. The columns are labeled, “Slopes,” “Intercepts,” “Type of Lines,” “Number of Solutions.” Under “Slopes” are “Different,” “Same,” and “Same.” Under “Intercepts,” the first cell is blank, then the words “Different” and “Same” appear. Under “Types of Lines” are the words, “Intersecting,” “Parallel,” and “Coincident.” Under “Number of Solutions” are “1 point,” “No Solution,” and “Infinitely many solutions.”

Let’s take one more look at our equations in (Example 7) that gave us parallel lines.

\left\{\begin{array}{c}y=\frac{1}{2}x-3\hfill \\ x-2y=4\hfill \end{array}

When both lines were in slope-intercept form we had:

y=\frac{1}{2}x-3\phantom{\rule{2em}{0ex}}y=\frac{1}{2}x-2

We call a system of equations like this an inconsistent system . It has no solution.

A system of equations that has at least one solution is called a consistent system .

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent equations , they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations , we get coincident lines.

Independent and Dependent Equations

Two equations are independent if they have different solutions.

Two equations are dependent if all the solutions of one equation are also solutions of the other equation.

Let’s sum this up by looking at the graphs of the three types of systems. See (Figure 3) and (Table 3) .

This figure shows three x y coordinate planes in a horizontal row. The first shows two lines intersecting. The second shows two parallel lines. The third shows two coincident lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

Without graphing, determine the number of solutions and then classify the system of equations.

\left\{\begin{array}{c}y=-2x-4\hfill \\ 4x+2y=9\hfill \end{array}

no solution, inconsistent, independent

\left\{\begin{array}{c}2x+y=-3\hfill \\ x-5y=5\hfill \end{array}.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

\left\{\begin{array}{c}3x+2y=2\hfill \\ 2x+y=1\hfill \end{array}

one solution, consistent, independent

\left\{\begin{array}{c}3x-2y=4\hfill \\ y=\frac{3}{2}x-2\hfill \end{array}

A system of equations whose graphs are coincident lines has infinitely many solutions and is consistent and dependent.

\left\{\begin{array}{c}4x-5y=20\hfill \\ y=\frac{4}{5}x-4\hfill \end{array}

infinitely many solutions, consistent, dependent

Solve Applications of Systems of Equations by Graphing

We will modify the  problem solving strategy slightly to set up and solve applications of systems of linear equations.

How to use a problem solving strategy for systems of linear equations.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

Step 5 is where we will use the method introduced in this section. We will graph the equations and find the solution.

Sondra is making 10 quarts of punch from fruit juice and club soda. The number of quarts of fruit juice is 4 times the number of quarts of club soda. How many quarts of fruit juice and how many quarts of club soda does Sondra need?

Step 1. Read the problem.

Step 2. Identify what we are looking for.

We are looking for the number of quarts of fruit juice and the number of quarts of club soda that Sondra will need.

Step 3. Name what we are looking for. Choose variables to represent those quantities.

f=

Step 4. Translate into a system of equations.

This figure shows sentences converted into equations. The first sentence reads, “The number of quarts of fruit juice and the number of quarts of club soda is 10. “Number of quarts of fruit juice” contains a curly bracket beneath the phrase with an “f” centered under the bracket. The “And” also contains a curly bracket beneath it and has a plus sign centered beneath it. “Number of quarts of club soda” contains a curly bracket with the variable “c” beneath it. And finally, the phrase “is 10” contains a curly bracket. Under this it reads equals 10. The second sentence reads, “The number of quarts of fruit juice is four times the number of quarts of club soda”. This sentence is set up similarly in that each phrase contains a curly bracket underneath. The variable “f” represents “The number of quarts of fruit juice”. An equal sign represents “is” and “4c” represents four times the number of quarts of club soda.”

Step 5. Solve the system of equations using good algebra techniques.

This figure shows two equations and their graph. The first equation is f = 4c where b = 4 and b = 0. The second equation is f + c = 10. f = negative c +10 where b = negative 1 and b = 10. The x y coordinate plane shows a graph of these two lines which intersect at (2, 8).

The point of intersection (2, 8) is the solution. This means Sondra needs 2 quarts of club soda and 8 quarts of fruit juice.

Step 6. Check the answer in the problem and make sure it makes sense.

Does this make sense in the problem?

Yes, the number of quarts of fruit juice, 8 is 4 times the number of quarts of club soda, 2.

Yes, 10 quarts of punch is 8 quarts of fruit juice plus 2 quarts of club soda.

Step 7. Answer the question with a complete sentence.

Sondra needs 8 quarts of fruit juice and 2 quarts of soda.

Manu is making 12 quarts of orange juice from concentrate and water. The number of quarts of water is 3 times the number of quarts of concentrate. How many quarts of concentrate and how many quarts of water does Manu need?

Manu needs 3 quarts juice concentrate and 9 quarts water.

Access these online resources for additional instruction and practice with solving systems of equations by graphing.

  • Instructional Video Solving Linear Systems by Graphing
  • Instructional Video Solve by Graphing

Key Concepts

4.1 exercise set.

In the following exercises, determine if the following points are solutions to the given system of equations.

\left\{\begin{array}{c}2x-6y=0\hfill \\ 3x-4y=5\hfill \end{array}

  • (\left(\frac{5}{4},\frac{7}{4}\right)\)

In the following exercises, solve the following systems of equations by graphing.

\left\{\begin{array}{c}3x+y=-3\hfill \\ 2x+3y=5\hfill \end{array}

Without graphing the following systems of equations, determine the number of solutions and then classify the system of equations.

\left\{\begin{array}{c}y=\frac{2}{3}x+1\hfill \\ -2x+3y=5\hfill \end{array}

In the following exercises, solve.

  • Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?
  • Enrique is making a party mix that contains raisins and nuts. For each ounce of nuts, he uses twice the amount of raisins. How many ounces of nuts and how many ounces of raisins does he need to make 24 ounces of party mix?
  • Leo is planning his spring flower garden. He wants to plant tulip and daffodil bulbs. He will plant 6 times as many daffodil bulbs as tulip bulbs. If he wants to plant 350 bulbs, how many tulip bulbs and how many daffodil bulbs should he plant?
  • a) no b) yes

\left(-2,3\right)

  • no solutions
  • no solutions, inconsistent, independent
  • consistent, 1 solution
  • Molly needs 16 ounces of strawberry juice and 48 ounces of water.
  • Enrique needs 8 ounces of nuts and 16 ounces of water.
  • Leo should plant 50 tulips and 300 daffodils.

Business/Technical Mathematics by Lynn Marecek and MaryAnne Anthony-Smith is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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  • How do you solve a system of equations by substitution?
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4.1: System of Equations - Graphing

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  • Darlene Diaz
  • Santiago Canyon College via ASCCC Open Educational Resources Initiative

Three Types of Solutions to a Linear System by Graphing

Given two linear equations and after graphing the lines,

Solution 1. If the two lines intersect, then the point of intersection is the solution to the system, i.e., the solution is an ordered-pair \((x, y)\).

Solution 2. If the two lines do not intersect and are parallel, i.e., they have the same slope and different \(y\)-intercepts, then the system has no solution .

Solution 3. If the two lines are the same line, then the solution is infinitely many solutions on that line.

Verifying Solutions

Example 4.1.1.

Is the ordered-pair \((2, 1)\) a solution to the system \[\left\{\begin{array}{l}3x-y=5 \\ x+y=3\end{array}\right. \quad ?\nonumber\]

To verify whether \((2, 1)\) is the solution to the system, we plug-n-chug \((2, 1)\) into each equation and determine whether we obtain a true statement. If we obtain true statements for both equations in the system, then \((2, 1)\) will be the solution to the system.

\[\begin{array}{rl}3x-y=5&\text{Plug-n-chug }x=2\text{ and }y=1 \\ 3(\color{blue}{2}\color{black}{})-(\color{blue}{1}\color{black}{})\stackrel{?}{=}5&\text{Simplify} \\ 6-1\stackrel{?}{=}5&\text{Subtract} \\ 5=5&\checkmark\:\text{True}\end{array}\nonumber\]

Let’s do the same for the second equation:

\[\begin{array}{rl}x+y=3&\text{Plug-n-chug }x=2\text{ and }y=1 \\ (\color{blue}{2}\color{black}{})+(\color{blue}{1}\color{black}{})\stackrel{?}{=}3&\text{Add} \\ 3=3&\checkmark\:\text{True}\end{array}\nonumber\]

Since the ordered-pair \((2, 1)\) makes both statements true, then \((2, 1)\) is a solution to the system. Hence, if we were to graph these lines, they would intersect at the point \((2, 1)\).

Example 4.1.2

Is the ordered-pair \((−3, −4)\) a solution to the system \[\left\{\begin{array}{l}5x+4y=-31 \\ 3x+6y=-36\end{array}\right.\quad ?\nonumber\]

To verify whether \((−3, −4)\) is the solution to the system, we plug-n-chug \((−3, −4)\) into each equation and determine whether we obtain a true statement. If we obtain true statements for both equations in the system, then \((−3, −4)\) will be the solution to the system.

\[\begin{array}{rl}5x+4y=-31&\text{Plug-n-chug }x=-3\text{ and }y=-4 \\ 5(\color{blue}{-3}\color{black}{})+4(\color{blue}{-4}\color{black}{})\stackrel{?}{=}-31&\text{Simplify} \\ -15-16\stackrel{?}{=}=31&\text{Subtract} \\ -31=-31&\checkmark\:\text{True}\end{array}\nonumber\]

\[\begin{array}{rl}3x+6y=-36&\text{Plug-n-chug }x=-3\text{ and }y=-4 \\ 3(\color{blue}{-3}\color{black}{})+6(\color{blue}{-4}\color{black}{})\stackrel{?}{=}-36&\text{Simplify} \\ -9-24\stackrel{?}{=}-36&\text{Subtract} \\ -33\neq -36&X \:\text{False}\end{array}\nonumber\]

Since the ordered-pair \((−3, −4)\) makes only one of the statements true, then \((−3, −4)\) is not a solution to the system. Recall, the ordered-pair must make the statement true for both equations in order to be a solution to the system.

Solve a System by Graphing

Example 4.1.3.

Solve the system by graphing: \[\left\{ \begin{array}{l}y=-\dfrac{1}{2}x+3 \\ y=\dfrac{3}{4}x-2\end{array}\right.\nonumber\]

We first need to decide the method in which we will graph. We learned in the previous chapter to make a table, use intercepts, or use the slope-intercept form. Notice both equations are given in slope-intercept form. Let’s go ahead and use the slope-intercept form to graph the lines.

\[\begin{array}{ll}y=-\dfrac{1}{2}x+3&(1) \\ y=\dfrac{3}{4}x-2&(2)\end{array}\nonumber\]

We will graph line (1) with a solid line and graph line (2) with a dashed line.

clipboard_e37fdd0cde29222f9cca9e9636ebc3c0c.png

We can see after graphing the two lines that they intersect at the point \((4, 1)\). Hence, the solution to the system is \((4, 1)\).

Example 4.1.4

Solve the system by graphing: \[\left\{\begin{array}{l}6x-3y=-9 \\ 2x+2y=-6\end{array}\right.\nonumber\]

We first need to decide the method in which we will graph. We learned in the previous chapter to make a table, use intercepts, or use the slope-intercept form. Since both equations are not given in slope-intercept form as in Example 4.1.3 , we can rewrite them in slope-intercept form, then graph. So, let’s rewrite each equation in slope-intercept form: \[\begin{array}{ll}6x-3y=-9 &2x+2y=-6 \\ -3y=-6x-9&2y=-2x-6 \\ y=\dfrac{-6x}{-3}-\dfrac{9}{-3}&y=\dfrac{-2}{2}x-\dfrac{6}{2} \\ y=2x+3&y=-x-3\end{array}\nonumber\]

Let’s go ahead and use the slope-intercept form to graph the lines.

\[\begin{array}{ll}y=2x+3&(1) \\ y=-x-3&(2)\end{array}\nonumber\]

clipboard_eac5f68886e164e569690b6e17df7ce43.png

We can see after graphing the two lines that they intersect at the point \((−2, −1)\). Hence, the solution to the system is \((−2, −1)\).

Example 4.1.5

Solve the system by graphing: \[\left\{\begin{array}{l}y=\dfrac{3}{2}x-4 \\ y=\dfrac{3}{2}x+1\end{array}\right.\nonumber\]

\[\begin{array}{ll}y=\dfrac{3}{2}x-4&(1) \\ y=\dfrac{3}{2}x+1&(2)\end{array}\nonumber\]

clipboard_ed4ecc8ec549ff84dd2732d5c57fa6940.png

We can see after graphing the two lines that these two lines are parallel. Hence, there is no solution to the system (since they will never intersect). Note, we could see by the system that these lines shared the same slope, but had different \(y\)-intercepts. Without graphing, we could have seen that these lines were parallel, hence, having no solution.

Example 4.1.6

Solve the system by graphing: \[\left\{\begin{array}{l}2x-6y=12 \\ 3x-9y=18\end{array}\right.\nonumber\]

We first need to decide the method in which we will graph. We learned in the previous chapter to make a table, use intercepts, or use the slope-intercept form. Since neither of the equations are written in slope-intercept form, let’s try graphing by making a table for each equation. Start with equation (1):

\[\begin{array}{lr} x=-3\quad & 2\color{blue}{(-3)}\color{black}{}-6y=12 \\ & -6-6y=12 \\ &-6y=18 \\ & y=-3\end{array}\nonumber\]

\[\begin{array}{lr} x=0\quad & 2\color{blue}{(0)}\color{black}{}-6y=12 \\ &-6y=12 \\ &y=-2\end{array}\nonumber\]

\[\begin{array}{lr} x=3\quad & 2\color{blue}{(3)}\color{black}{}-6y=12 \\ &6-6y=12 \\ &-6y=6 \\ &y=-1 \end{array}\nonumber\]

Next, equation (2):

\[\begin{array}{lr} x=-3\quad &3\color{blue}{(-3)}\color{black}{}-9y=18 \\ &-9-9y=18 \\ &-9y=27 \\ &y=-3\end{array}\nonumber\]

\[\begin{array}{lr} x=0\quad &3\color{blue}{(0)}\color{black}{}-9y=18 \\ &-9y=18 \\ &y=-2\end{array}\nonumber\]

\[\begin{array}{lr}x=3\quad &3\color{blue}{(3)}\color{black}{}-9y=18 \\ &9-9y=18 \\ &-9y=9 \\ &y=-1\end{array}\nonumber\]

Now, let’s graph the ordered-pairs. We will graph line (1) with a solid line and graph line (2) with a dashed line.

clipboard_e4e3cfec76ed78801aa0e6a9ca03941a0.png

We can see after graphing the two lines that these two lines are the same. Hence, there are infinitely many solutions on the line \(2x − 6y = 12\) (or the other equation) to the system (since they intersect at every point on the line). Note, we could see by the system that these lines shared the same slope and \(y\)-intercepts (after putting them in slope-intercept form). Without graphing, we could have seen that these lines were the same line, hence, having infinitely many solutions.

The Babylonians were the first to work with systems of equations with two variables. However, their work with systems was quickly passed by the Greeks, around 300 AD, who would solve systems of equations with three or four variables and eventually developed methods for solving systems with any number of unknowns.

Systems of Equations: Graphing Homework

Determine whether the given ordered pair(s) is a solution to the system.

Exercise 4.1.1

\(\begin{array}{l}2x+8y=0 \\ -8x+3y=38;\: (-4,1)\end{array}\)

Exercise 4.1.2

\(\begin{array}{l}-5x+6y=11 \\ 4x+2y=-4;\: (-1,1)\end{array}\)

Exercise 4.1.3

\(\begin{array}{l}6x+5y=49 \\ -x-6y=-34;\: (4,5)\end{array}\)

Exercise 4.1.4

\(\begin{array}{l} -2x+2y=-8 \\ -6x-3y=-6;\: (2,-2)\end{array}\)

Solve each system by graphing.

Exercise 4.1.5

\(\begin{array}{l}y=-x+1 \\ y=-5x-3\end{array}\)

Exercise 4.1.6

\(\begin{array}{l}y=-3 \\ y=-x-4\end{array}\)

Exercise 4.1.7

\(\begin{array}{l}y=-\dfrac{3}{4}x+1 \\ y=-\dfrac{3}{4}x+2\end{array}\)

Exercise 4.1.8

\(\begin{array}{l}y=\dfrac{1}{3}x+2 \\ y=-\dfrac{5}{3}x-4\end{array}\)

Exercise 4.1.9

\(\begin{array}{l}y=\dfrac{5}{3}x+4 \\ y=-\dfrac{2}{3}x-3\end{array}\)

Exercise 4.1.10

\(\begin{array}{l}x+3y=-9 \\ 5x+3y=3\end{array}\)

Exercise 4.1.11

\(\begin{array}{l}x-y=4 \\ 2x+y=-1\end{array}\)

Exercise 4.1.12

\(\begin{array}{l}2x+3y=-6 \\ 2x+y=2\end{array}\)

Exercise 4.1.13

\(\begin{array}{l}2x+y=2 \\ x-y=4\end{array}\)

Exercise 4.1.14

\(\begin{array}{l}2x+y=-2 \\ x+3y=9\end{array}\)

Exercise 4.1.15

\(\begin{array}{l}0=-6x-9y+36 \\ 12=6x-3y\end{array}\)

Exercise 4.1.16

\(\begin{array}{l}2x-y=-1 \\ 0=-2x-y-3\end{array}\)

Exercise 4.1.17

\(\begin{array}{l}3+y=-x \\ -4-6x=-y\end{array}\)

Exercise 4.1.18

\(\begin{array}{l}-y+7x=4 \\ -y-3+7x=0\end{array}\)

Exercise 4.1.19

\(\begin{array}{l}-12+x=4y \\ 12-5x=4y\end{array}\)

Exercise 4.1.20

\(\begin{array}{l}y=-\dfrac{5}{4}x-2 \\ y=-\dfrac{1}{4}x+2\end{array}\)

Exercise 4.1.21

\(\begin{array}{l}y=-x-2 \\ y=\dfrac{2}{3}x+3\end{array}\)

Exercise 4.1.22

\(\begin{array}{l}y=2x+2 \\ y=-x-4\end{array}\)

Exercise 4.1.23

\(\begin{array}{l}y=2x-4 \\ y=\dfrac{1}{2}x+2\end{array}\)

Exercise 4.1.24

\(\begin{array}{l}y=\dfrac{1}{2}x+4 \\ y=\dfrac{1}{2}x+1\end{array}\)

Exercise 4.1.25

\(\begin{array}{l}x+4y=-12 \\ 2x+y=4\end{array}\)

Exercise 4.1.26

\(\begin{array}{l}6x+y=-3 \\ x+y=2\end{array}\)

Exercise 4.1.27

\(\begin{array}{l}3x+2y=2 \\ 3x+2y=-6\end{array}\)

Exercise 4.1.28

\(\begin{array}{l}x+2y=6 \\ 5x-4y=16\end{array}\)

Exercise 4.1.29

\(\begin{array}{l}x-y=3 \\ 5x+2y=8\end{array}\)

Exercise 4.1.30

\(\begin{array}{l}-2y+x=4 \\ 2=-x+\dfrac{1}{2}y\end{array}\)

Exercise 4.1.31

\(\begin{array}{l}-2y=-4-x \\ -2y=-5x+4\end{array}\)

Exercise 4.1.32

\(\begin{array}{l}16=-x-4y \\ -2x=-4-4y\end{array}\)

Exercise 4.1.33

\(\begin{array}{l}-4+y=x \\ x+2=-y\end{array}\)

Exercise 4.1.34

\(\begin{array}{l}-5x+1=-y \\ -y+x=-3\end{array}\)

IMAGES

  1. Solving Systems By Graphing Worksheet

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  3. Unit 5 Lesson 3 Homework (Solving Systems by Graphing #2)

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  4. Solving Linear Systems By Graphing Worksheet

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  5. Unit 11 Lesson 1: Solving Systems by Graphing

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  6. How to Solve Systems of Equations by Graphing

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  3. McGraw Hill Module6 Lesson 1 Solving Systems of Equations by Graphing

  4. Algebra 1 STAAR Practice Solving Systems of Equations (A.5C

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  1. 5.1: Solve Systems of Equations by Graphing

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    -1- ©y n2M0E1N2x VKQumt6aX xSxo6f MtNwuarhe 0 bLTLjC e.D g gA ql0l e XroiNguh9t Msn lr ceyspeTrhv4e Md5.L 3 WMPaOd EeZ AwFift Xh6 HIQnMf1i qnOi Btfe 3 MAGlLg9e hb Dr9aI H1R.3 Worksheet by Kuta Software LLC

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  4. 4.1: Solving Linear Systems by Graphing

    The graphing method for solving linear systems is not ideal when the solution consists of coordinates that are not integers. There will be more accurate algebraic methods in sections to come, but for now, the goal is to understand the geometry involved when solving systems. It is important to remember that the solutions to a system correspond ...

  5. 4.1: Solving Systems by Graphing

    Reporting your solution on your homework: In reporting your solution on your homework paper, follow the Calculator Submission Guidelines from Chapter 3, Section 2. Make an accurate copy of the image shown in your viewing window. ... This page titled 4.1: Solving Systems by Graphing is shared under a CC BY-NC-ND 3.0 license and was authored, ...

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    © 2024 Google LLC Get this full course at http://www.MathTutorDVD.comIn this lesson we will learn about systems of equations and how to solve them using the method of graphing...

  7. PDF 8.1 SOLVING SYSTEMS BY GRAPHING AND SUBSTITUTION

    First write the equations in slope-intercept form: a l c u l a t o r l o s e - u p To check Example 1,graph y1 x 2 and y2 x 4. From the CALC menu, choose intersect to have the calcu-lator locate the point of intersection of the two lines.

  8. Systems of equations with graphing (video)

    When solving systems of linear equations, one method is to graph both equations on the same coordinate plane. The intersection of the two lines represents a solution that satisfies both equations. Other, more mathematical, methods may also be used. Created by Sal Khan. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted

  9. 5.1 Solve Systems of Equations by Graphing

    Try It 5.16. Solve each system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16. If you write the second equation in Example 5.8 in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept. When we graphed the second line in the last example, we drew it right over the first line.

  10. How to Solve Systems of Equations by Graphing

    Step 1: Analyze what form each equation of the system is in. Step 2: Graph the equations using the slope and y-intercept or using the x- and y-intercepts. Case 1: If the equations are in the slope-intercept form, identify the slope and y-intercept and graph them.

  11. 4.1 Solving Systems by Graphing

    A2.5.4 Solve systems of linear equations and inequalities in two variables by substitution, graphing, and use matrices with three variables; Section 4.1 Solving Systems Graphically ... a2_4.1_packet.pdf: File Size: 1138 kb: File Type: pdf: Download File. Practice Solutions. a2_4.1_solutions.pdf: File Size: 421 kb: File Type: pdf: Download File ...

  12. 4.1 Solve Systems of Linear Equations with Two Variables

    Try It 4.6. Solve the system by graphing: { 2 x + y = 6 x + y = 1. In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we'll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

  13. Quiz 1: Solving Linear Systems by Graphing Flashcards

    linear inequality. an open sentence of the form Ax + By + C < 0 or Ax + By + C > 0. The point will lie in the shaded region of both graphs if it is... true. The point will lie outside of the shaded region of both graphs if it is... false. In order for an ordered pair to be a solution of a system, it must satisfy...

  14. Solving Systems of Equations by Graphing Worksheets

    This is called "solving by graphing" and is a valid approach for linear equations with relatively simple slope and y-intercept values. The graphing systems of equations worksheets on this page fit this criteria and they are good practice for building a visual intuition of the solution process. In practice, it the linear equations in a system ...

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    To solve a system of linear equations by graphing. Graph the first equation. Graph the second equation on the same rectangular coordinate system. Determine whether the lines intersect, are parallel, or are the same line. Identify the solution to the system. If the lines intersect, identify the point of intersection.

  16. PDF 7.1 Solving Systems of Equations by Graphing

    Example 1: a. Is (-1, 0) a solution to ? b. Is (-1, 7) a solution to ? Solution: To determine if the ordered pair is a solution, we need to determine if it satisfies each of the equations in the system. To do that, we simply plug the ordered pair in and see if we get a true statement. ( ) ( ) ( ) ( )

  17. Chapter 3 Section 1 Homework

    Homework help video over how to solve a system of Linear Equations by graphing. Also, how to classify a system based on its number of solutions. All example...

  18. Homework 1: Solving Systems by Graphing

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  19. 3.1: Solve Systems of Linear Equations with Two Variables

    A solution of a system of two linear equations is represented by an ordered pair (x, y). To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system. Example 3.1.1.

  20. 4.1.0 Solving Systems of Linear Equations (Graphing)

    Solving Systems of Equations - Graphing 1. Teacher 15 terms. Mary_Conant3. Preview. Terms in this set (20) (-4, -1) Solve by graphing (-1, 1) Solve by graphing (1, -2) Solve by graphing (4, -2) y = 1/2x - 4 y = -3x + 10 (4, 2) Find the solution. (2, 3) Find the solution. (3, 2) Find the solution. (0, 0) Find the solution. (2, 4)

  21. System of Equations Calculator

    How do you solve a system of equations by graphing? To solve a system of equations by graphing, graph both equations on the same set of axes and find the points at which the graphs intersect. Those points are the solutions. How do you solve a system of equations by elimination?

  22. 5.1: Solve Systems of Equations by Graphing

    Solve each system by graphing: {−x + y = 1 2x + y = 10 { − x + y = 1 2 x + y = 10. Answer. Exercise 5.1.12 5.1. 12. Solve each system by graphing: {2x + y = 6 x + y = 1 { 2 x + y = 6 x + y = 1. Answer. Usually when equations are given in standard form, the most convenient way to graph them is by using the intercepts.

  23. 4.1: System of Equations

    Solution 1. If the two lines intersect, then the point of intersection is the solution to the system, i.e., the solution is an ordered-pair (x, y). Solution 2. If the two lines do not intersect and are parallel, i.e., they have the same slope and different y -intercepts, then the system has no solution. Solution 3.

  24. Answered: Use a software program or a graphing…

    Use a software program or a graphing utility with matrix capabilities and Cramer's Rule to solve (if possible) the system of linear equations. (If not possible, enter IMPOSSIBLE.) -8x₁ + 7x2 10x3 = -155 12x₁ + 3x25x3 = -9 15x₁9x2 + 2x3 = 153 (X₁₁ X2₁ x 3) =