ORCCA Substitution

Section 5.2 Substitution

In Section 5.1, we focused on solving systems of equations by graphing. In addition to being time consuming, graphing can be an awkward method to determine the exact solution when the solution has large numbers or fractions. There are two symbolic methods for solving systems of linear equations, and in this section we will use one of them: substitution.

Subsection 5.2.1 Solving Systems of Equations Using Substitution

Example 5.2.1. The Interview.

In 2014, the New York Times¹ posted the following about the movie, “The Interview” :

(nyti.ms/2pupebT)

“The Interview” generated roughly $\$15$ million in online sales and rentals during its first four days of availability, Sony Pictures said on Sunday.

Sony did not say how much of that total represented $\$6$ digital rentals versus $\$15$ sales. The studio said there were about two million transactions overall.

A few days later, Joey Devilla cleverly pointed out in his blog², that there is enough information given to find the amount of sales versus rentals. Using algebra, we can write a system of equations and solve it to find the two quantities.³

http://www.joeydevilla.com/2014/12/31/

Although since the given information uses approximate values, the solutions we will find will only be approximations too.

First, we will define variables. We need two variables, because there are two unknown quantities: how many sales there were and how many rentals there were. Let $r$ be the number of rental transactions and let $s$ be the number of sales transactions.

If you are unsure how to write an equation from the background information, use the units to help you. The units of each term in an equation must match because we can only add like quantities. Both $r$ and $s$ are in transactions. The article says that the total number of transactions is $2$ million. So our first equation will add the total number of rental and sales transactions and set that equal to $2$ million. Our equation is:

\begin{equation*} (r\,\text{transactions})+(s\,\text{transactions})=2{,}000{,}000\,\text{transactions} \end{equation*}

Without the units:

\begin{equation*} r+s=2{,}000{,}000 \end{equation*}

The price of each rental was $\$6\text{.}$ That means the problem has given us a rate of $6\,\frac{\text{dollars}}{\text{transaction}}$ to work with. The rate unit suggests this should be multiplied by something measured in transactions. It makes sense to multiply by $r\text{,}$ and then the number of dollars generated from rentals was $6r\text{.}$ Similarly, the price of each sale was $\$15\text{,}$ so the revenue from sales was $15s\text{.}$ The total revenue was $\$15$ million, which we can represent with this equation:

\begin{equation*} \left(6\,\tfrac{\text{dollars}}{\text{transaction}}\right)(r\,\text{transactions})+\left(15\,\tfrac{\text{dollars}}{\text{transaction}}\right)(s\,\text{transactions})=\$15{,}000{,}000 \end{equation*}

Without the units:

\begin{equation*} 6r+15s=15{,}000{,}000 \end{equation*}

Here is our system of equations:

\begin{equation*} \left\{ \begin{alignedat}{4} r\amp+{}\amp s\amp={}\amp2{,}000{,}000 \\ 6r\amp+{}\amp 15s\amp={}\amp15{,}000{,}000 \end{alignedat} \right. \end{equation*}

To solve the system, we will use the substitution method. The idea is to use one equation to find an expression that is equal to $r$ but, cleverly, does not use the variable “$r\text{.}$” Then, substitute this for $r$ into the other equation. This leaves you with one equation that only has one variable.

The first equation from the system is an easy one to solve for $r\text{:}$

\begin{align*} r+s \amp=2{,}000{,}000\\ r \amp=2{,}000{,}000-s \end{align*}

This tells us that the expression $2{,}000{,}000-s$ is equal to $r\text{,}$ so we can substitute it for $r$ in the second equation:

\begin{align*} 6r+15s \amp=15{,}000{,}000\\ 6(\substitute{2{,}000{,}000-s})+15s \amp=15{,}000{,}000\\ \end{align*}

Now we have an equation with only one variable, $s\text{,}$ which we will solve for:

\begin{align*} 6(2{,}000{,}000-s)+15s \amp=15{,}000{,}000\\ 12{,}000{,}000-6s+15s \amp=15{,}000{,}000\\ 12{,}000{,}000+9s \amp= 15{,}000{,}000\\ 9s \amp= 3{,}000{,}000\\ \divideunder{9s}{9} \amp= \divideunder{3{,}000{,}000}{9}\\ s \amp= 333{,}333.\overline{3} \end{align*}

At this point, we know that $s=333{,}333.\overline{3}\text{.}$ This tells us that out of the $2$ million transactions, roughly $333{,}333$ were from online sales. Recall that we solved the first equation for $r\text{,}$ and found $r=2{,}000{,}000-s\text{.}$

\begin{align*} r \amp=2{,}000{,}000-s\\ r \amp=2{,}000{,}000-\substitute{333{,}333.\overline{3}}\\ r \amp=1{,}666{,}666.\overline{6} \end{align*}

To check our answer, we will see if $s=333{,}333.\overline{3}$ and $r=1{,}666{,}666.\overline{6}$ make the original equations true:

\begin{align*} r+s \amp=2{,}000{,}000\\ \substitute{1{,}666{,}666.\overline{6}}+\substitute{333{,}333.\overline{3}} \amp\stackrel{?}{=}2{,}000{,}000\\ 2{,}000{,}000\amp\stackrel{\checkmark}{=}2{,}000{,}000 \end{align*}

\begin{align*} 6r+15s \amp=15{,}000{,}000\\ 6\left(\substitute{1{,}666{,}666.\overline{6}}\right)+15\left(\substitute{333{,}333.\overline{3}}\right) \amp\stackrel{?}{=}15{,}000{,}000\\ 10{,}000{,}000+5{,}000{,}000 \amp\stackrel{\checkmark}{=}15{,}000{,}000 \end{align*}

In summary, there were roughly $333{,}333$ copies sold and roughly $1{,}666{,}667$ copies rented.

Remark 5.2.2.

In Example 5.2.1, we chose to solve the equation $r+s=2{,}000{,}000$ for $r\text{.}$ We could just as easily have instead solved for $s$ and substituted that result into the second equation instead. The summary conclusion would have been the same.

Remark 5.2.3.

In Example 5.2.1, we rounded the solution values because only whole numbers make sense in the context of the problem. It was OK to round, because the original information we had to work with were rounded. In fact, it would be OK to round even more to $s=330{,}000$ and $r=1{,}700{,}000\text{,}$ as long as we communicate clearly that we rounded and our values are rough.

In other exercises where there is no context and nothing suggests the given numbers are approximations, it is not OK to round and all answers should be communicated with their exact values.

Example 5.2.4.

Solve the system of equations using substitution:

\begin{align*} \left\{ \begin{alignedat}{4} x\amp+{}\amp 2y \amp={}\amp 8 \\ 3x\amp-{}\amp 2y \amp={}\amp 8 \\ \end{alignedat} \right. \end{align*}

Explanation

To use substitution, we need to solve for one of the variables in one of our equations. Looking at both equations, it will be easiest to solve for $x$ in the first equation:

\begin{align*} x+2y\amp=8\\ x\amp= 8-2y \end{align*}

Next, we replace $x$ in the second equation with $8-2y\text{,}$ giving us a linear equation in only one variable, $y\text{,}$ that we may solve:

\begin{align*} 3x-2y\amp=8\\ 3(\substitute{8-2y})-2y\amp=8\\ 24-6y-2y\amp=8\\ 24-8y\amp=8\\ -8y\amp=-16\\ y\amp=2 \end{align*}

Now that we have the value for $y\text{,}$ we need to find the value for $x\text{.}$ We have already solved the first equation for $x\text{,}$ so that is the easiest equation to use.

\begin{align*} x\amp= 8-2y\\ x\amp= 8-2(\substitute{2})\\ x\amp= 8-4\\ x\amp=4 \end{align*}

To check this solution, we replace $x$ with $4$ and $y$ with $2$ in each equation:

\begin{align*} x+2y\amp=8\amp3x-2y\amp=8\\ \substitute{4}+2\substitute{(2)}\amp\stackrel{?}{=}8\amp3\substitute{(4)}-2\substitute{(2)}\amp\stackrel{?}{=}8\\ 4+4\amp\stackrel{\checkmark}{=}8\amp12-4\amp\stackrel{\checkmark}{=}8 \end{align*}

We conclude then that this system of equations is true when $x=4$ and $y=2\text{.}$ Our solution is the point $(4,2)$ and we write the solution set as $\{(4,2)\}\text{.}$

Example 5.2.5.

Solve this system of equations using substitution:

\begin{align*} \left\{ \begin{alignedat}{4} 3x\amp-{}\amp 7y \amp={}\amp 5 \\ -5x\amp+{}\amp 2y \amp={}\amp 11 \\ \end{alignedat} \right. \end{align*}

Explanation

We need to solve for one of the variables in one of our equations. Looking at both equations, it will be easiest to solve for $y$ in the second equation. The coefficient of $y$ in that equation is smallest.

\begin{align*} -5x+2y\amp=11\\ 2y\amp=11+5x\\ \divideunder{2y}{2}\amp=\divideunder{11+5x}{2}\\ y\amp=\frac{11}{2}+\frac{5}{2}x \end{align*}

Note that in this example, there are fractions once we solve for $y\text{.}$ We should take care with the steps that follow that the fraction arithmetic is correct.

Replace $y$ in the first equation with $\frac{11}{2}+\frac{5}{2}x\text{,}$ giving us a linear equation in only one variable, $x\text{,}$ that we may solve:

\begin{align*} 3x-7y\amp=5\\ 3x-7\left(\frac{11}{2}+\frac{5}{2}x\right)\amp=5\\ 3x-7\cdot\frac{11}{2}-7\cdot\frac{5}{2}x\amp=5\\ 3x-\frac{77}{2}-\frac{35}{2}x\amp=5\\ \frac{6}{2}x-\frac{77}{2}-\frac{35}{2}x\amp=5\\ -\frac{29}{2}x-\frac{77}{2}\amp=5\\ -\frac{29}{2}x\amp=\frac{10}{2}\addright{\frac{77}{2}}\\ -\frac{29}{2}x\amp=\frac{87}{2}\\ \multiplyleft{-\frac{2}{29}}\left(-\frac{29}{2}x\right)\amp=\multiplyleft{-\frac{2}{29}}\left(\frac{87}{2}\right)\\ x\amp=-3 \end{align*}

Now that we have the value for $x\text{,}$ we need to find the value for $y\text{.}$ We have already solved the second equation for $y\text{,}$ so that is the easiest equation to use.

\begin{align*} y\amp=\frac{11}{2}+\frac{5}{2}x\\ y\amp=\frac{11}{2}+\frac{5}{2}(\substitute{-3})\\ y\amp=\frac{11}{2}-\frac{15}{2}\\ y\amp=-\frac{4}{2}\\ y\amp=-2 \end{align*}

To check this solution, we replace $x$ with $-3$ and $y$ with $-2$ in each equation:

\begin{align*} 3x-7y \amp=5 \amp -5x+2y \amp= 11\\ 3(-3)-7(-2) \amp\stackrel{?}{=}5 \amp -5(-3)+2(-2) \amp\stackrel{?}{=} 11\\ -9+14\amp\stackrel{\checkmark}{=}5\amp 15-4\amp\stackrel{\checkmark}{=}11 \end{align*}

We conclude then that this system of equations is true when $x=-3$ and $y=-2\text{.}$ Our solution is the point $(-3,-2)$ and we write the solution set as $\{(-3,-2)\}\text{.}$

Example 5.2.6. Clearing Fraction Denominators Before Solving.

Solve the system of equations using the substitution method:

\begin{align*} \left\{ \begin{aligned} \frac{x}{3} - \frac{1}{2}y \amp= \frac{5}{6} \\ \frac{1}{4}x \amp = \frac{y}{2} + 1 \\ \end{aligned} \right. \end{align*}

Explanation

When a system of equations has fraction coefficients, it can be helpful to take steps that replace the fractions with whole numbers. With each equation, we may multiply each side by the least common denominator of all the fractions.

In the first equation, the least common denominator is $6\text{,}$ so:

\begin{align*} \frac{x}{3}-\frac{1}{2}y\amp=\frac{5}{6}\\ \multiplyleft{6}\left(\frac{x}{3}-\frac{1}{2}y\right)\amp=\multiplyleft{6}\frac{5}{6}\\ 6\cdot\frac{x}{3}-6\cdot\frac{1}{2}y\amp=\multiplyleft{6}\frac{5}{6}\\ 2x-3y\amp=5 \end{align*}

In the second equation, the least common denominator is $4\text{,}$ so:

\begin{align*} \frac{1}{4}x\amp=\frac{y}{2}+1\\ \multiplyleft{4}\frac{1}{4}x\amp=\multiplyleft{4}\frac{y}{2}+\multiplyleft{4}1\\ \multiplyleft{4}\frac{1}{4}x\amp=\multiplyleft{4}\frac{y}{2}+\multiplyleft{4}1\\ x\amp=2y+4 \end{align*}

Now we have this system that is equivalent to the original system of equations, but there are no fraction coefficients:

\begin{align*} \left\{ \begin{aligned} 2x-3y\amp=5 \\ x\amp=2y+4 \\ \end{aligned} \right. \end{align*}

The second equation is already solved for $x\text{,}$ so we will substitute $x$ in the first equation with $2y+4\text{,}$ and we have:

\begin{align*} 2x-3y\amp=5\\ 2(\substitute{2y+4})-3y\amp=5\\ 4y+8-3y\amp=5\\ y+8\amp=5\\ y\amp=-3 \end{align*}

And we have solved for $y\text{.}$ To find $x\text{,}$ we know $x=2y+4\text{,}$ so we have:

\begin{align*} x\amp=2y+4\\ x\amp=2(\substitute{-3})+4\\ x\amp=-6+4\\ x\amp=-2 \end{align*}

The solution is $(-2,-3)\text{.}$ Checking this solution is left as an exercise.

Checkpoint 5.2.7.

Try a similar exercise.

Subsection 5.2.2 Solving Special Systems of Equations with Substitution

Remember the two special cases we encountered when solving by graphing in Section 5.1? If the two lines represented by a system of equations have the same slope, then they might be separate lines that never meet, meaning the system has no solutions. Or they might coincide as the same line, in which case there are infinitely many solutions represented by all the points on that line. Let's see what happens when we use the substitution method on each of the special cases.

Example 5.2.8. A System with No Solution.

Solve the system of equations using the substitution method:

\begin{align*} \left\{ \begin{aligned} y \amp= 2x-1 \\ 4x - 2y \amp= 3 \\ \end{aligned} \right. \end{align*}

Explanation

Since the first equation is already solved for $y\text{,}$ we will substitute $2x-1$ for $y$ in the second equation, and we have:

\begin{align*} 4x-2y\amp=3\\ 4x-2\substitute{(2x-1)}\amp=3\\ 4x-4x+2\amp=3\\ 2\amp=3 \end{align*}

Even though we were only intending to substitute away $y\text{,}$ we ended up with an equation where there are no variables at all. This will happen whenever the lines have the same slope. This tells us the system represents either parallel or coinciding lines. Since $2=3$ is false no matter what values $x$ and $y$ might be, there can be no solution to the system. So the lines are parallel and distinct. We write the solution set using the empty set symbol: the solution set is $\emptyset\text{.}$

To verify this, re-write the second equation, $4x-2y=3\text{,}$ in slope-intercept form:

\begin{align*} 4x-2y\amp=3\\ -2y\amp=-4x+3\\ \divideunder{-2y}{-2}\amp=\divideunder{-4x+3}{-2}\\ y\amp=\frac{-4x}{-2}+\frac{3}{-2}\\ y\amp=2x-\frac{3}{2} \end{align*}

So the system is equivalent to:

\begin{align*} \left\{ \begin{aligned} y \amp = 2x-1 \\ y \amp = 2x-\frac{3}{2} \\ \end{aligned} \right. \end{align*}

Now it is easier to see that the two lines have the same slope, but different $y$-intercepts. They are parallel and distinct lines, so the system has no solution.

Example 5.2.9. A System with Infinitely Many Solutions.

Solve the system of equations using the substitution method:

\begin{align*} \left\{ \begin{aligned} y \amp=2x-1 \\ 4x-2y \amp=2 \\ \end{aligned} \right. \end{align*}

Explanation

Since $y=2x-1\text{,}$ we will substitute $2x-1$ for $y$ in the second equation and we have:

\begin{align*} 4x-2y\amp=2\\ 4x-2\substitute{(2x-1)}\amp=2\\ 4x-4x+2\amp=2\\ 2\amp=2 \end{align*}

Even though we were only intending to substitute away $y\text{,}$ we ended up with an equation where there are no variables at all. This will happen whenever the lines have the same slope. This tells us the system represents either parallel or coinciding lines. Since $2=2$ is true no matter what values $x$ and $y$ might be, the system equations are true no matter what $x$ is, as long as $y=2x-1\text{.}$ So the lines coincide. We write the solution set as $\{(x,y)\mid y=2x-1\}\text{.}$

To verify this, re-write the second equation, $4x-2y=2\text{,}$ in slope-intercept form:

\begin{align*} 4x-2y\amp=2\\ -2y\amp=-4x+2\\ \frac{-2y}{-2}\amp=\frac{-4x}{-2}+\frac{2}{-2}\\ y\amp=2x-1 \end{align*}

The system looks like:

\begin{align*} \left\{ \begin{alignedat}{4} y \amp {}={} \amp 2x-1 \\ y \amp {}={} \amp 2x-1 \\ \end{alignedat} \right. \end{align*}

Now it is easier to see that the two equations represent the same line. Every point on the line is a solution to the system, so the system has infinitely many solutions.

Exercises 5.2.3 Exercises

Review and Warmup

1.

Solve the equation.

$\displaystyle{ {{\frac{3}{2}}-5t}={10} }$

2.

Solve the equation.

$\displaystyle{ {{\frac{7}{8}}-2b}={10} }$

3.

Solve the equation.

$\displaystyle{ {{\frac{3}{4}}-{\frac{1}{4}}A}={9} }$

4.

Solve the equation.

$\displaystyle{ {{\frac{5}{2}}-{\frac{1}{2}}B}={6} }$

5.

Solve the linear equation for $y\text{.}$

\begin{equation*} {28x+4y}={-40} \end{equation*}

6.

Solve the linear equation for $y\text{.}$

\begin{equation*} {25x-5y}={-90} \end{equation*}

7.

Solve the linear equation for $y\text{.}$

\begin{equation*} {15x+5y}={35} \end{equation*}

8.

Solve the linear equation for $y\text{.}$

\begin{equation*} {2y-16x}={-6} \end{equation*}

9.

Solve the linear equation for $y\text{.}$

\begin{equation*} {6x-y}={8} \end{equation*}

10.

Solve the linear equation for $y\text{.}$

\begin{equation*} {4x-y}={-17} \end{equation*}