ORCCA Substitution

Section 4.2 Substitution

Objectives: PCC Course Content and Outcome Guide

MTH 60 CCOG 7.1
MTH 60 CCOG 7.2

In Section 4.1, 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, fractions, or decimals. There are two symbolic methods for solving systems of linear equations, and in this section we will use one of them: substitution.

Figure 4.2.1. Alternative Video Lesson

Subsection 4.2.1 Solving Systems of Equations Using Substitution

Example 4.2.2. The Interview.

In 2014, the New York Times¹(nyti.ms/2pupebT) posted the following about the movie, “The Interview”:

“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 blog²http://www.joeydevilla.com/2014/12/31/, 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.³Although 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\amp 6r+15s \amp=15{,}000{,}000\\ \substitute{1{,}666{,}666.\overline{6}}+\substitute{333{,}333.\overline{3}} \amp\stackrel{?}{=}2{,}000{,}000\amp 6\left(\substitute{1{,}666{,}666.\overline{6}}\right)+15\left(\substitute{333{,}333.\overline{3}}\right) \amp\stackrel{?}{=}15{,}000{,}000\\ 2{,}000{,}000\amp\stackrel{\checkmark}{=}2{,}000{,}000\amp 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 4.2.3.

In Example 4.2.2, 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 4.2.4.

In Example 4.2.2, 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 4.2.5.

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 $\{(4,2)\}\text{.}$

Checkpoint 4.2.6.

Example 4.2.7.

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 4.2.8. 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 multiple of all the denominators.

In the first equation, the least common multiple of the denominators 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 multiple of the denominators 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 4.2.9.

Try a similar exercise.

For summary reference, here is the general procedure.

Process 4.2.10. Solving Systems of Equations by Substitution.

To solve a system of equations by substitution,

Solve one of the equations for one of the variables.
Substitute that expression into the other equation. There should now only be one variable in that equation.
Solve that equation for the one remaining variable.
Substitute that value into an earlier equation and solve for the other variable.
Verify your solution in the original equations.

Subsection 4.2.2 Applications of Systems of Equations

Example 4.2.11. Two Different Interest Rates.

Notah made some large purchases with his two credit cards one month and took on a total of $\$8{,}400$ in debt from the two cards. He didn't make any payments the first month, so the two credit card debts each started to accrue interest. That month, his Visa card charged $2\%$ interest and his Mastercard charged $2.5\%$ interest. Because of this, Notah's total debt grew by $\$178\text{.}$ How much money did Notah charge to each card?

Explanation

To start, we will define two variables based on our two unknowns. Let $v$ be the amount charged to the Visa card (in dollars) and let $m$ be the amount charged to the Mastercard (in dollars).

To determine our equations, notice that we are given two different totals. We will use these to form our two equations. The total amount charged is $\$8{,}400$ so we have:

\begin{equation*} (v\,\text{dollars})+(m\,\text{dollars})=\$8400 \end{equation*}

Or without units:

\begin{equation*} v+m=8400 \end{equation*}

The other total we were given is the total amount of interest, $\$178\text{,}$ which is also in dollars. The Visa had $v$ dollars charged to it and accrues $2\%$ interest. So $0.02v$ is the dollar amount of interest that comes from using this card. Similarly, $0.025m$ is the dollar amount of interest from using the Mastercard. Together:

\begin{equation*} 0.02(v\,\text{dollars})+0.025(m\,\text{dollars})=\$178 \end{equation*}

Or without units:

\begin{equation*} 0.02v+0.025m=178 \end{equation*}

As a system, we write:

\begin{equation*} \left\{ \begin{alignedat}{3} v\amp+{}\amp m\amp={}8400 \\ 0.02v\amp+{}\amp 0.025m\amp{}=178 \\ \end{alignedat} \right. \end{equation*}

To solve this system by substitution, notice that it will be easier to solve for one of the variables in the first equation. We'll solve that equation for $v\text{:}$

\begin{align*} v+m\amp=8400\\ v\amp=8400-m \end{align*}

Now we will substitute $8400-m$ for $v$ in the second equation:

\begin{align*} 0.02v+0.025m\amp=178\\ 0.02(\substitute{8400-m})+0.025m\amp=178\\ 168-0.02m+0.025m\amp=178\\ 168+0.005m\amp=178\\ 0.005m\amp=10\\ \divideunder{0.005m}{0.005}\amp=\divideunder{10}{0.005}\\ m\amp=2000 \end{align*}

Lastly, we can determine the value of $v$ by using the earlier equation where we isolated $v\text{:}$

\begin{align*} v\amp=8400-m\\ v\amp=8400-\substitute{2000}\\ v\amp=6400 \end{align*}

In summary, Notah charged $\$6400$ to the Visa and $\$2000$ to the Mastercard. We should check that these numbers work as solutions to our original system and that they make sense in context. (For instance, if one of these numbers were negative, or was something small like $\$0.50\text{,}$ they wouldn't make sense as credit card debt.)

The next two examples are called mixture problems, because they involve mixing two quantities together to form a combination and we want to find out how much of each quantity to mix.

Example 4.2.12. Mixing Solutions with Two Different Concentrations.

LaVonda is a meticulous bartender and she needs to serve $600$ milliliters of Rob Roy, an alcoholic cocktail that is $34\%$ alcohol by volume. The main ingredients are scotch that is $42\%$ alcohol and vermouth that is $18\%$ alcohol. How many milliliters of each ingredient should she mix together to make the concentration she needs?

Explanation

The two unknowns are the quantities of each ingredient. Let $s$ be the amount of scotch (in mL) and let $v$ be the amount of vermouth (in mL).

One quantity given to us in the problem is 600 mL. Since this is the total volume of the mixed drink, we must have:

\begin{equation*} (s\,\text{mL})+(v\,\text{mL})=600\,\text{mL} \end{equation*}

Or without units:

\begin{equation*} s+v=600 \end{equation*}

To build the second equation, we have to think about the alcohol concentrations for the scotch, vermouth, and Rob Roy. It can be tricky to think about percentages like these correctly. One strategy is to focus on the amount (in mL) of alcohol being mixed. If we have $s$ milliliters of scotch that is $42\%$ alcohol, then $0.42s$ is the actual amount (in mL) of alcohol in that scotch. Similarly, $0.18v$ is the amount of alcohol in the vermouth. And the final cocktail is 600 mL of liquid that is $34\%$ alcohol, so it has $0.34(600)=204$ milliliters of alcohol. All this means:

\begin{equation*} 0.42(s\,\text{mL})+0.18(v\,\text{mL})=204\,\text{mL} \end{equation*}

Or without units:

\begin{equation*} 0.42s+0.18v=204 \end{equation*}

So our system is:

\begin{equation*} \left\{ \begin{alignedat}{3} s\amp+{}\amp v\amp=600 \\ 0.42s\amp+{}\amp0.18v\amp=204 \end{alignedat} \right. \end{equation*}

To solve this system, we'll solve for $s$ in the first equation:

\begin{align*} s+v\amp=600\\ s\amp=600-v \end{align*}

And then substitute $s$ in the second equation with $600-v\text{:}$

\begin{align*} 0.42s+0.18v\amp=204\\ 0.42(\substitute{600-v})+0.18v\amp=204\\ 252-0.42v+0.18v\amp=204\\ 252-0.24v\amp=204\\ -0.24v\amp=-48\\ \divideunder{-0.24v}{-0.24}\amp=\divideunder{-48}{-0.24}\\ v\amp=200 \end{align*}

As a last step, we will determine $s$ using the equation where we had isolated $s\text{:}$

\begin{align*} s\amp=600-v\\ s\amp=600-\substitute{200}\\ s\amp=400 \end{align*}

In summary, LaVonda needs to combine 400 mL of scotch with 200 mL of vermouth to create 600 mL of Rob Roy that is $34\%$ alcohol by volume.

As a check for Example 4.2.12, we will use estimation to see that our solution is reasonable. Since LaVonda is making a $34\%$ solution, she would need to use more of the $42\%$ concentration than the $18\%$ concentration, because $34\%$ is closer to $42\%$ than to $18\%\text{.}$ This agrees with our answer because we found that she needed 400 mL of the $42\%$ solution and 200 mL of the $18\%$ solution. This is an added check that we have found reasonable answers.

Example 4.2.13. Mixing a Coffee Blend.

Desi owns a coffee shop and they want to mix two different types of coffee beans to make a blend that sells for $\$12.50$ per pound. They have some coffee beans from Columbia that sell for $\$9.00$ per pound and some coffee beans from Honduras that sell for $\$14.00$ per pound. How many pounds of each should they mix to make $30$ pounds of the blend?

Explanation

Before we begin, it may be helpful to try to estimate the solution. Let's compare the three prices. Since $\$12.50$ is between the prices of $\$9.00$ and $\$14.00\text{,}$ this mixture is possible. Now we need to estimate the amount of each type needed. The price of the blend ($\$12.50$ per pound) is closer to the higher priced beans ($\$14.00$ per pound) than the lower priced beans ($\$9.00$ per pound). So we will need to use more of that type. Keeping in mind that we need a total of $30$ pounds, we roughly estimate $20$ pounds of the $\$14.00$ Honduran beans and $10$ pounds of the $\$9.00$ Columbian beans. How good is our estimate? Next we will solve this exercise exactly.

To set up our system of equations we define variables, letting $C$ be the amount of Columbian coffee beans (in pounds) and $H$ be the amount of Honduran coffee beans (in pounds).

The equations in our system will come from the total amount of beans and the total cost. The equation for the total amount of beans can be written as:

\begin{equation*} (C\,\text{lb})+(H\,\text{lb})=30\,\text{lb} \end{equation*}

Or without units:

\begin{equation*} C+H=30 \end{equation*}

To build the second equation, we have to think about the cost of all these beans. If we have $C$ pounds of Columbian beans that cost $\$9.00$ per pound, then $9C$ is the cost of those beans in dollars. Similarly, $14H$ is the cost of the Honduran beans. And the total cost is for $30$ pounds of beans priced at $\$12.50$ per pound, totaling $12.5(30)=37.5$ dollars. All this means:

\begin{equation*} \left(9\,\tfrac{\text{dollars}}{\text{lb}}\right)(C\,\text{lb})+\left(14\,\tfrac{\text{dollars}}{\text{lb}}\right)(H\,\text{lb})=\left(12.50\,\tfrac{\text{dollars}}{\text{lb}}\right)(30\,\text{lb}) \end{equation*}

Or without units and carrying out the multiplication on the right:

\begin{equation*} 9C+14H=37.5 \end{equation*}

Now our system is:

\begin{equation*} \left\{ \begin{alignedat}{3} C\amp+{}\amp H\amp=30 \\ 9C\amp+{}\amp 14H\amp=37.50 \end{alignedat} \right. \end{equation*}

To solve the system, we'll solve the first equation for $C\text{:}$

\begin{align*} C+H\amp=30\\ C\amp=30-H \end{align*}

Next, we'll substitute $C$ in the second equation with $30-H\text{:}$

\begin{align*} 9C+14H\amp=375\\ 9(\substitute{30-H})+14H\amp=375\\ 270-9H+14H\amp=375\\ 270+5H\amp=375\\ 5H\amp=105\\ H\amp=21 \end{align*}

Since $H=21\text{,}$ we can conclude that $C=9\text{.}$

In summary, Desi needs to mix $21$ pounds of the Honduran coffee beans with $9$ pounds of the Columbian coffee beans to create this blend. Our estimate at the beginning was pretty close, so we feel this answer is reasonable.

Subsection 4.2.3 Solving Special Systems of Equations with Substitution

Remember the two special cases we encountered when solving by graphing in Subsection 4.1.2? 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 4.2.14. 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 4.2.15. 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. The solution set is $\{(x,y) \mid y=2x-1\}\text{.}$

Reading Questions 4.2.4 Reading Questions

1.

Give an example of a system of two equations in $x$ and $y$ where it would be nicer to solve the system using substitution than by graphing the two lines that the equations define. Explain why substitution would be nicer than graphing for your example system.

2.

What might be a good first step if you have a system of two linear equations in two variables where there are fractions appearing in the equations?

3.

In an application problem, thinking about the can help you understand how to set up equations.

Exercises 4.2.5 Exercises

Review and Warmup

1.

Solve the equation.

$\displaystyle{ {{\frac{5}{8}}-9a}={6} }$

2.

Solve the equation.

$\displaystyle{ {{\frac{7}{4}}-7c}={6} }$

3.

Solve the equation.

$\displaystyle{ {{\frac{3}{10}}-{\frac{1}{10}}A}={4} }$

4.

Solve the equation.

$\displaystyle{ {{\frac{7}{8}}-{\frac{1}{8}}C}={1} }$

5.

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

${-40x-5y}={-50}$

6.

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

${5y-15x}={-5}$

7.

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

${25x+5y}={-60}$

8.

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

${12x-2y}={-34}$

9.

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

${6x-y}={-16}$

10.

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

${3x-y}={-2}$