High School Algebra I Unlocked (2016)

GOALS

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

•Define and categorize systems of linear equations

•Solve systems of equations questions with methods such as graphing, substitution, addition or subtraction, and algebra

•Calculate the slopes of parallel and perpendicular lines

•Graph and solve linear inequalities

Lesson 5.1. Systems of Equations

In Chapter 4, we learned how to solve and graph linear equations, which appear as lines in the coordinate plane. But what happens if more than one line exists in a coordinate plane? Whenever you are dealing with multiple equations at the same time, you are working with a system of equations, which are also referred to as simultaneous equations, or a two-variable system of equations. A system of equations is a set of equations that share variables and work together in a coordinate plane. The solution(s) to a system of equations will be the point(s) of intersection.

Look back to Lesson 4.1
for more on linear equations.
In this chapter, we’re
talking about systems of
equations. A system is a
set of connected things
or parts that form a
complex whole. So, when
we talk about a system of
equations, we are talking
about a set of equations
that work together.

A linear two-variable system of equations can have an infinite number of solutions, one solution, or no solution. The solution to a system of equations is usually expressed in the form (x, y), where x represents the x-coordinate, and yrepresents the y-coordinate at the point of intersection.

When a system of equations has either an infinite number of solutions or exactly one solution, the equations are referred to as consistent. When a system of equations has no solution, the equations are referred to as inconsistent. Consistent systems are categorized as either dependent or independent. A system of equations with exactly one solution is known as an independent system, as each equation provides unique information. Conversely, a system of equations with an infinite number of solutions is known as a dependent system because the equations provide identical information.

There are multiple ways to solve a system of equations problem—four, to be exact.

Four Methods for Solving a System of Equations

1. Graphing Method

2. Substitution Method

3. Addition/Subtraction Method

4. Algebraic Method

First, let’s take a look at the graphical approach.

Solving Systems of Equations Graphically

EXAMPLE

y = 2x + 10

y = x + 8

Solve the system of linear equations by graphing.

Unlike the questions we tackled in Chapter 4, this question gives you two equations and asks you to solve the system. Start by graphing each equation, as you would graph any other linear equation, in a coordinate plane:

When working with a single linear equation, the solution is any point that falls on the line. However, when working with a system of equations, the solution is the point of intersection. So refer to the graph and find the point where the two lines meet.

As you can see, the solution to this system of equations is x = −2 and y = 6. If you want to check your work, try plugging the values x = −2 and y = 6 back into the original equations. If both sides of the equation are equal to one another, the solution is correct. Take a look at how this works:

Since both equations hold true with the values x = −2 and y = 6, you know that your solution is correct. And that’s it: You’ve successfully conquered your first system of equations problem! The equations in this system are consistent, and the solution is (−2, 6).

You may also encounter questions that require you to determine whether or not a given point is a solution to a system of equations. In this scenario, you will plug the coordinate into the equations to see if the equation holds true; this is similar to the process used to check your solution in Example 1.

EXAMPLE

y = 4x − 6

y = −4x + 2

Which of the following points is a solution to the above system of equations?

A) (1, 2)

B) (−1, 2)

C) (1, −2)

D) (−1, −2)

Unlike Example 1, which asked you to find the solution to the system of equations by graphing, this question requires you to identify the solution to the system of equations, or the point at which the equations intersect (the point of intersection). In order to determine which point is the solution, plug the x- and y-values into the equations. If both sides of the equation are equal to one another, the coordinate is the point of intersection.

Let’s try this out. Start with (A), and plug x = 1 and y = 2 into the first equation:

y = 4x − 6

2 = 4(1) − 6

2 = 4 − 6

2 = −2

Since 2 ≠ −2, the coordinate (1, 2) is not a point of intersection and, therefore, is not the solution to the system of equations. Now plug (B), x = −1 and y = 2, into the first equation:

y = 4x − 6

2 = 4(−1) − 6

2 = −4 − 6

2 = −10

Here, 2 ≠ −10, so the coordinate (−1, 2) is not a solution to the system of equations. Now let’s tackle (C), plugging x = 1 and y = −2 into the first equation:

y = 4x − 6

−2 = 4(1) − 6

−2 = 4 − 6

−2 = −2

Here, −2 = −2, so this is a solution to the first equation and could be a solution to the system of equations; keep (C). Finally, plug (D), x = −1 and y = −2, into the first equation:

y = 4x − 6

−2 = 4(−1) − 6

−2 = −4 − 6

−2 = −10

Since −2 ≠ −10, the point (−1, −2) is not a point of intersection, and (D) can be eliminated. Therefore, the only point of intersection for this system of equations is (1, −2), and (C) is the correct answer. If more than one choice had been a solution to the first equation, then you would determine which is a solution to the system of equations by plugging the choices you didn’t eliminate into the second equation. Only one choice will work for both.

That was a lengthy process, so you may be wondering if there’s an alternative way of determining the point of intersection. The answer is yes! Just set the two equations equal to each other, solve for x, and then substitute the x-value into one of the original equations to solve for y. Let’s see how this works.

4x − 6 = −4x + 2

4x + 4x = 2 + 6 

8x = 8

x = 1

Now that you have found x = 1, plug this value into one of the original equations to solve for y:

y = −4x + 2

y = −4(1) + 2

y = −2

Therefore, the point of intersection is (1, −2). To check your answer, you can graph the lines and find the point of intersection. This will result in the following graph that shows the point of intersection at (1, −2), confirming that our answer is indeed correct.

Regardless of the method you choose to use to find the point of intersection—plugging in the values given in the answer choices, setting the equations equal to one another, or graphing the set of equations—you will get the same answer!

Now that you have this process down, let’s try another question.

EXAMPLE

2y = 18x + 12

−3 = 4.5x − y

Solve the following system of equations graphically.

Up until this point, the equations that we’ve dealt with have all been in slope-intercept form, which makes graphing quite straightforward. Here, neither equation is in slope-intercept form, so you must rewrite each equation in the form y = mx + b. Let’s start by rewriting and graphing the first equation.

2y = 18x + 12

y = 9x + 6

Now that the equation is in slope-intercept form, y = mx + b, you can identify the slope, m, and the y-intercept, b. Therefore, the slope of this equation is 9, the y-intercept occurs at (0, 6), and you can graph the line as follows:

Next, rewrite the second equation in slope-intercept form:

−3 = 4.5x − y

−3 = x −y

−6 = 9x − y

−9x − 6 = −y

y = 9x + 6

Since the equation is in slope-intercept form, y = mx + b, you can now identify the slope, m, and the y-intercept, b. Here the slope of the equation is 9, the y-intercept occurs at (0, 6), and you can graph the line:

Did you notice that the equations are identical? When the two equations are the same, it means that the lines are the same and that the system of equations has an infinite number of solutions; the equations are consistent and dependent.

Thus, the solution to this system of equations is y = 9x + 6. If you were asked to write this solution as a point on the coordinate plane, the solution would be (x, 9x + 6), indicating that for every value of x, the associated value of y is 9x + 6.

Solving Systems of Equations by Substitution

Solving a system of equations graphically or by determining whether the two lines are equal to one another may not always be feasible or preferable. Another method for solving a system of equations is by substitution.

The substitution method involves solving one equation for one variable, plugging that value into the other equation, and then solving for the second variable. Then, take the second variable and substitute it back into the first equation to determine the value of the first variable. This might sound like a complicated process, so let’s see how it works in practice.

EXAMPLE

9x = 3y + 18

2 + 4x = y

Solve the following system of equations by substitution.

Unlike the previous examples, you do not need to rewrite the equations in slope-intercept form when using the substitution method. Rather, you simply need to either isolate x or y in a single equation. In this scenario, the second equation is already in slope-intercept form, and the value of y is 2 + 4x.

First, substitute y = 2 + 4x into the first equation, 9x = 3y + 18, and solve for x as follows:

9x = 3y + 18

9x = 3(2 + 4x) + 18

9x = 6 + 12x + 18

9x − 12x = 6 + 18

−3x = 24

x = −8

Next, substitute the value x = −8 into the second equation, 2 + 4x = y, as follows:

y = 2 + 4x

y = 2 + 4(−8)

y = 2 − 32

y = −30

The solution to this system of equations is (−8, −30).

If you want to check your work and make sure that your solution set holds true, plug x = −8 and y = −30 into either equation. This will result in the following

Since both equations result in true statements, we can be confident that the solution to this system of equations is, in fact, (−8, −30).

So, how would the substitution method work when the lines in a system of equations are identical? Let’s look at an example of this type of scenario.

EXAMPLE

6y = 6x + 24

−7x = −7y + 28

Solve the following system of equations by substitution.

To use the substitution method, start by isolating either the x- or y-value in either equation. Let’s begin by isolating y in the first equation:

6y = 6x + 24

y = x + 4

Now substitute the value y = x + 4 into the second equation to solve for x as follows:

−7x = −7y + 28

−7x = −7(x + 4) + 28

−7x = −7x − 28 + 28

−7x + 7x = −28 + 28

0 = 0

What happened here? We didn’t find a value for x! No worries. Whenever you use the substitution method and find that you are unable to solve for the second variable—instead finding that both sides of the equation are equal to one another—you are dealing with a set of consistent, dependent equations. This means that the equations are identical to one another, there is no point of intersection, and there is an infinite number of solutions to this system of equations. Therefore, the solution to this system of equations is (x, x + 4), indicating that for every value of x, the associated value of y is x + 4.

Any time you solve a system of equations and find a statement that says a numerical value is equal to a numerical value, such as 0 = 0, 8 = 8, or 1,000 = 1,000, you are dealing with a dependent system of equations. In these situations, the system of equations has an infinite number of solutions.

Solving Systems of Equations by Addition or Subtraction

In addition to the graphical and substitution methods of solving a system of equations, you can use the addition or subtraction method, which is also sometimes referred to as the elimination method. The addition, or elimination, method works in the same way that it does when you’re working with a single linear equation. For example, if we were working with the single linear equation 4x + 12 = 16, we would use the addition method to simplify the equation as follows:

When working with a system of equations, you will use the same strategy. Let’s see how this works.

EXAMPLE

6x + 4y = 8

4x − 4y = 12

Solve the following system of equations using the addition method.

You’ll want to use the addition method for solving a system of equations when either the coefficient attached to one variable is identical in both equations, or when the equations can be manipulated to make the coefficients equivalent; this will make adding the equations straightforward. In this situation, both equations have 4y as a term, so you can add the equations as follows:

Once you’ve isolated the x variable, solve for x:

10x = 20

x = 2

Now plug x = 2 into either of the equations; regardless of the equation you choose, the result will be the same. Here’s how you can find the y-value in each equation:

Thus, the solution to the system of equations is (2, −1). To check your work, plug x = 2 and y = −1 back into either of the original equations.

Since both of the equations hold true when x = 2 and y = −1, you can be confident that your solution, (2, −1), is correct.

Note: Although we’ve been plugging these values back into the equations back into the original equations to check our work, this is not a mandatory step in the solving process. Rather, this process of back-solving provides assurance that you found the correct solution set, so we highly recommend you include this step when solving problems.

Let’s try another question using the addition method.

EXAMPLE

3y − 4x = 7

12y − 24x = 12

If the solution to the following system of equations is (x, y), what is the value of x + y ?

If you noticed that none of the coefficients match, don’t panic; see if the equations can be manipulated to create terms that will cancel when the equations are added together. Here, the second equation can be manipulated:

12y − 24x = 12→ − E₂ → −3y + 6x = −3

The notation used in between the arrows, − E₂, indicates that you are multiplying the second equation by −. Now that you have a revised second equation, you can add the equations:

Now that you’ve isolated the x variable, solve for x:

2x = 4

x = 2

Next, plug in x = 2 into either of the equations. Regardless of the equation you choose, the result will be the same. Take a look at how you can find the y-value in each of the original equations:

Thus, the solution to the system of equations is (2, 5). Another way you can check your work is by graphing these equations and identifying the point of intersection as follows:

Here is how you may see systems of equations on the SAT.

y = 3x − 1

y + x = 1

In the system of equations above, if (x, y) is the solution to the system, what is the value of ?

A)

B)

C)

D)

As you can see in the graph of the two original equations, the lines intersect at the point (2, 5).

However, we aren’t quite done with this question. Note that the question asks you to find the value of x + y. Since the solution to our system of equations is (2, 5), the value of x + y = 2 + 5 = 7.

Solving Systems of Equations with Algebra

There is another way to solve simultaneous equations that relies on—drumroll, please—algebra! The algebraic approach to solving a system of equations involves setting the two equations equal to one another, solving for one variable, substituting the found value into one of the found equations, and solving for the second variable. Let’s see how this works.

EXAMPLE

y = 20x − 4

−y = 5x + 29

Solve the following system of equations algebraically.

To use the algebraic method, start by isolating either the x- or y-value in both equations. Since the y-value is already isolated in the second equation, start by manipulating the second equation to make the y-value positive.

−y = 5x + 29

y = −5x − 29

Since both of the equations represent the value of y, you can set the two equations equal to one another and solve for x:

 20x − 4 = −5x − 29

20x + 5x = 4 − 29

25x = −25

 x = −1

Now that you’ve found x = −1, plug this value into either equation to solve for y:

As with the other methods we covered, you will find an identical y-value regardless of the equation you use to solve for y. The solution to this system of equations is (−1, −24).

Let’s try another example using the algebraic approach.

EXAMPLE

8y + 24x = −32

12x − 6y + 18 = 0

Solve the following system of equations algebraically and find the value of (2y + x)².

Because neither variable is isolated in the given simultaneous equations, you need to simplify and isolate one of the variables. Here, start by isolating the y-variable.

While you can use any method to solve this question, let’s use the addition method. The first step is to clear the fraction in the second equation by multiplying all terms by 2:

1/2y + x = 1 → 2 E₂ → y + 2x = 2

Now stack and subtract the two equations to solve for x:

 x = 3/5

Now that you’ve found the value of x = 3/5, plug this value into the first equation to solve for y:

 y − 3x = −1

y − 3(3/5) = −1

y − 9/5 = −1

    y = −5/5 + 9/5

  y = 4/5

Great! You now know that x = 3/5 and y = 4/5. However, the question asks you to find the value of x/y, so set up the fraction and simplify:

Thus, the value of x/y = 3/4, or (C).

Now that both of the equations represent the value of y, set the two equations equal to one another and solve for x:

−3x − 4 = 2x + 3

5x = −7

 x = −

Now, plug x = − into either equation to solve for y:

As before, regardless of the equation you use, the y-value will be identical. Thus, the solution to this system of equations is (−, ). However, we aren’t done yet! The question asks us to find the value of (2y + x)². So, plug in the values of x and y values as follows:

Now you’re done. The value of (2y + x)² is 1.

Here is how you may see systems of equations on the ACT.

If a + 3b = 27 and a − 3b = 9, then b = ?

F. 3

G. 9

H. 14

J. 18

K. 36

Now that you can handle all types of simultaneous equations, let’s discuss some special relationships that can exist between two linear equations.