# 7.2: Systems Using Substitution

**At Grade**Created by: CK-12

**Practice**Systems Using Substitution

What if you were given a system of linear equations like \begin{align*}x - y = 7\end{align*} and \begin{align*}3x - 4y = -3\end{align*}? How could you substitute one equation into the other to solve for the variables? After completing this Concept, you'll be able to solve a system of linear equations by substitution.

### Try This

For lots more practice solving linear systems, check out this web page: http://www.algebra.com/algebra/homework/coordinate/practice-linear-system.epl

After clicking to see the solution to a problem, you can click the back button and then click Try Another Practice Linear System to see another problem.

### Guidance

In this lesson, we’ll learn to solve a system of two equations using the method of substitution.

**Solving Linear Systems Using Substitution of Variable Expressions**

Let’s look again at the problem about Peter and Nadia racing.

*Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?*

In that example we came up with two equations:

Nadia’s equation: \begin{align*}d = 6t\end{align*}

Peter’s equation: \begin{align*}d = 5t + 20\end{align*}

Each equation produced its own line on a graph, and to solve the system we found the point at which the lines intersected—the point where the values for \begin{align*}d\end{align*} and \begin{align*}t\end{align*} satisfied **both** relationships. When the values for \begin{align*}d\end{align*} and \begin{align*}t\end{align*} are equal, that means that Peter and Nadia are at the same place at the same time.

But there’s a faster way than graphing to solve this system of equations. Since we want the value of \begin{align*}d\end{align*} to be the same in both equations, we could just set the two right-hand sides of the equations equal to each other to solve for \begin{align*}t\end{align*}. That is, if \begin{align*}d = 6t\end{align*} and \begin{align*}d = 5t + 20\end{align*}, and the two \begin{align*}d\end{align*}’s are equal to each other, then by the transitive property we have \begin{align*}6t = 5t + 20\end{align*}. We can solve this for \begin{align*}t\end{align*}:

\begin{align*}6t &= 5t + 20 && subtract \ 5t \ from \ both \ sides:\\ t &= 20 && substitute \ this \ value \ for \ t \ into \ Nadia's \ equation:\\ d &= 6 \cdot 20 = 120\end{align*}

Even if the equations weren’t so obvious, we could use simple algebraic manipulation to find an expression for one variable in terms of the other. If we rearrange Peter’s equation to isolate \begin{align*}t\end{align*}:

\begin{align*}d &= 5t + 20 && subtract \ 20 \ from \ both \ sides:\\ d - 20 &= 5t && divide \ by \ 5:\\ \frac{d - 20}{5} &= t\end{align*}

We can now ** substitute** this expression for \begin{align*}t\end{align*} into Nadia’s equation \begin{align*}(d = 6t)\end{align*} to solve:

\begin{align*}d &= 6 \left ( \frac{d - 20}{5} \right ) && multiply \ both \ sides \ by \ 5:\\ 5d &= 6(d - 20) && distribute \ the \ 6:\\ 5d &= 6d - 120 && subtract \ 6d \ from \ both \ sides:\\ -d &= -120 && divide \ by \ -1:\\ d &= 120 && substitute \ value \ for \ d \ into \ our \ expression \ for \ t:\\ t &= \frac{120 - 20}{5} = \frac{100}{5} = 20\end{align*}

So we find that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 feet away.

The method we just used is called the **Substitution Method.** In this lesson you’ll learn several techniques for isolating variables in a system of equations, and for using those expressions to solve systems of equations that describe situations like this one.

#### Example A

Let’s look at an example where the equations are written in **standard form.**

*Solve the system*

\begin{align*}2x + 3y & = 6\\ -4x + y & = 2\end{align*}

Again, we start by looking to isolate one variable in either equation. If you look at the second equation, you should see that the coefficient of \begin{align*}y\end{align*} is 1. So the easiest way to start is to use this equation to solve for \begin{align*}y\end{align*}.

Solve the second equation for \begin{align*}y\end{align*}:

\begin{align*}-4x + y &= 2 && add \ 4x \ to \ both \ sides:\\ y &= 2 + 4x\end{align*}

Substitute this expression into the first equation:

\begin{align*}2x + 3(2 + 4x) &= 6 && distribute \ the \ 3:\\ 2x + 6 + 12x &= 6 && collect \ like \ terms:\\ 14x + 6 &= 6 && subtract \ 6 \ from \ both \ sides:\\ 14x &= 0 && and \ hence:\\ x &= 0\end{align*}

Substitute back into our expression for \begin{align*}y\end{align*}:

\begin{align*}y = 2 + 4 \cdot 0 = 2\end{align*}

As you can see, we end up with the same solution \begin{align*}(x = 0, y = 2)\end{align*} that we found when we graphed these functions back in Lesson 7.1. So long as you are careful with the algebra, the substitution method can be a very efficient way to solve systems.

Next, let’s look at a more complicated example. Here, the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} we end up with aren’t whole numbers, so they would be difficult to read off a graph!

#### Example B

*Solve the system*

\begin{align*}2x + 3y & = 3\\ 2x - 3y & = -1\end{align*}

Again, we start by looking to isolate one variable in either equation. In this case it doesn’t matter which equation we use—all the variables look about equally easy to solve for.

So let’s solve the first equation for \begin{align*}x\end{align*}:

\begin{align*}2x + 3y &= 3 && subtract \ 3y \ from \ both \ sides:\\ 2x &= 3 - 3y && divide \ both \ sides \ by \ 2:\\ x &= \frac{1}{2}(3 - 3y)\end{align*}

Substitute this expression into the second equation:

\begin{align*}\cancel{2} \cdot \frac{1}{2}(3 - 3y)-3y &= -1 && cancel \ the \ fraction \ and \ re-write \ terms:\\ 3 - 3y - 3y &= -1 && collect \ like \ terms:\\ 3 - 6y &= -1 && subtract \ 3 \ from \ both \ sides:\\ -6y &= -4 && divide \ by \ -6:\\ y &= \frac{2}{3}\end{align*}

Substitute into the expression we got for \begin{align*}x\end{align*}:

\begin{align*}x & = \frac{1}{2} \left ( 3 - \cancel{3} \left ( \frac{2}{\cancel{3}} \right ) \right )\\ x & = \frac{1}{2}\end{align*}

So our solution is \begin{align*}x = \frac{1}{2}, y = \frac{2}{3}\end{align*}. You can see how the graphical solution \begin{align*}\left ( \frac{1}{2}, \frac{2}{3} \right )\end{align*} might have been difficult to read accurately off a graph!

**Solving Real-World Problems Using Linear Systems**

Simultaneous equations can help us solve many real-world problems. We may be considering a purchase—for example, trying to decide whether it’s cheaper to buy an item online where you pay shipping or at the store where you do not. Or you may wish to join a CD music club, but aren’t sure if you would really save any money by buying a new CD every month in that way. Or you might be considering two different phone contracts. Let’s look at an example of that now.

#### Example C

*Anne is trying to choose between two phone plans. The first plan, with Vendafone, costs $20 per month, with calls costing an additional 25 cents per minute. The second company, Sellnet, charges $40 per month, but calls cost only 8 cents per minute. Which should she choose?*

You should see that Anne’s choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the *number of minutes* is the independent variable, it will be our \begin{align*}x\end{align*}. Cost is *dependent* on minutes – the *cost per month* is the dependent variable and will be assigned \begin{align*}y\end{align*}.

*For Vendafone:* \begin{align*}y = 0.25x + 20\end{align*}

*For Sellnet:* \begin{align*}y = 0.08x + 40\end{align*}

By writing the equations in slope-intercept form \begin{align*}(y = mx + b)\end{align*}, you can sketch a graph to visualize the situation:

The line for Vendafone has an intercept of 20 and a slope of 0.25. The Sellnet line has an intercept of 40 and a slope of 0.08 (which is roughly a third of the Vendafone line’s slope). In order to help Anne decide which to choose, we’ll find where the two lines cross, by solving the two equations as a system.

Since equation 1 gives us an expression for \begin{align*}y (0.25x + 20)\end{align*}, we can substitute this expression directly into equation 2:

\begin{align*}0.25x + 20 &= 0.08x + 40 && subtract \ 20 \ from \ both \ sides:\\ 0.25x &= 0.08x + 20 && subtract \ 0.08x \ from \ both \ sides:\\ 0.17x &= 20 && divide \ both \ sides \ by \ 0.17:\\ x &= 117.65 \ minutes && rounded \ to \ 2 \ decimal \ places.\end{align*}

So if Anne uses 117.65 minutes a month (although she can’t really do *exactly* that, because phone plans only count whole numbers of minutes), the phone plans will cost the same. Now we need to look at the graph to see which plan is better if she uses more minutes than that, and which plan is better if she uses fewer. You can see that the Vendafone plan costs more when she uses more minutes, and the Sellnet plan costs more with fewer minutes.

So, **if Anne will use 117 minutes or less every month she should choose** *Vendafone.***If she plans on using 118 or more minutes she should choose** *Sellnet.*

Watch this video for help with the Examples above.

CK-12 Foundation: Linear Systems by Substitution

### Guided Practice

*Solve the system*

\begin{align*}8x + 10y & = 2\\ 4x - 15y & = -19\end{align*}

**Solution:**

Again, we start by looking to isolate one variable in either equation. In this case it doesn’t matter which equation we use—all the variables look about equally easy to solve for.

So let’s solve the first equation for \begin{align*}x\end{align*}:

\begin{align*}8x + 10y & = 2 && subtract \ 10y \ from \ both \ sides:\\ 8x &= 2 - 10y && divide \ both \ sides \ by \ 8:\\ x &= \frac{1}{8}(2 - 10y)\end{align*}

Substitute this expression into the second equation:

\begin{align*}4 \cdot \frac{1}{8}(2 - 10y)-15y &= -19 && simplify \ the \ fraction:\\ \frac{1}{2}(2 - 10y)-15y &= -19 && distribute \ the \ fraction \ and \ re-write \ terms:\\ 1 - 5y - 15y &= -19 && collect \ like \ terms:\\ 1 - 20y &= -19 && subtract \ 1 \ from \ both \ sides:\\ -20y &= -20 && divide \ by \ -20:\\ y &= 1 \end{align*}

Substitute into the expression we got for \begin{align*}x\end{align*}:

\begin{align*}x &= \frac{1}{8}(2 - 10y) && Substitute \ the \ y-value \ into \ the \ x \ equation: \\ x &= \frac{1}{8}(2 - 10(1)) && Simplify: \\ x &= \frac{1}{8}(2 - 10) \\ x &= \frac{1}{8}(-8) \\ x & = -1\end{align*}

So our solution is \begin{align*}x = -1, y = 1\end{align*}.

### Explore More

- Solve the system: \begin{align*}x + 2y =9\!\\ 3x + 5y = 20\end{align*}
- Solve the system: \begin{align*}x - 3y =10\!\\ 2x + y = 13\end{align*}
- Solve the system: \begin{align*}2x + 0.5y = -10\!\\ x - y = -10\end{align*}
- Solve the system: \begin{align*}2x + 0.5y = 3\!\\ x + 2y = 8.5\end{align*}
- Solve the system: \begin{align*}3x + 5y = -1\!\\ x + 2y = -1\end{align*}
- Solve the system: \begin{align*}3x + 5y = -3\!\\ x + 2y = - \frac{4}{3}\end{align*}
- Solve the system: \begin{align*}x - y = - \frac{12}{5}\!\\ 2x + 5y = -2\end{align*}
- Of the two non-right angles in a right angled triangle, one measures twice as many degrees as the other. What are the angles?
- The sum of two numbers is 70. They differ by 11. What are the numbers?
- A number plus half of another number equals 6; twice the first number minus three times the second number equals 4. What are the numbers?
- A rectangular field is enclosed by a fence on three sides and a wall on the fourth side. The total length of the fence is 320 yards. If the field has a total perimeter of 400 yards, what are the dimensions of the field?
- A ray cuts a line forming two angles. The difference between the two angles is \begin{align*}18^\circ\end{align*}. What does each angle measure?
- Jason is five years older than Becky, and the sum of their ages is 23. What are their ages?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 7.2.

Consistent

A system of equations is consistent if it has at least one solution.distributive property

The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, .linear equation

A linear equation is an equation between two variables that produces a straight line when graphed.substitute

In algebra, to substitute means to replace a variable or term with a specific value.### Image Attributions

Here you'll learn how to use subtitution to solve systems of linear equations in two variables. You'll then solve real-world problems involving such systems.