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## Section5.4Systems of Linear Equations Chapter Review

### Subsection5.4.1Solving Systems of Linear Equations by Graphing

In Section 5.1 we covered the definition of system of linear equations and how a solution to a system of linear equation is a point where the graphs of the two equations cross. We also considered special systems of equations that overlap or never touch.

###### Example5.4.1Solving Systems of Linear Equations by Graphing

Solve the following system of equations by graphing:

\begin{equation*} \left\{ \begin{alignedat}{4} y \amp {}={} \amp {-\frac{2}{3}x} \amp {}-{} \amp 4 \\ y \amp {}={} \amp {-4x} \amp {}-{} \amp 14 \\ \end{alignedat} \right. \end{equation*}
Explanation

The first equation, $y=-\frac{2}{3}x-4\text{,}$ is a linear equation in slope-intercept form with a slope of $-\frac{2}{3}$ and a $y$-intercept of $(0,-4)\text{.}$ The second equation, $y=-4x-14\text{,}$ is a linear equation in slope-intercept form with a slope of $-4$ and a $y$-intercept of $(0,-14)\text{.}$ We'll use this information to graph both lines in Figure 5.4.2.

The two lines intersect where $x=-3$ and $y=-2\text{,}$ so the solution of the system of equations is the point $(-3,-2)\text{.}$ We write the solution set as $\{(-3,-2)\}\text{.}$ ###### Example5.4.3Special Systems of Equations

Solve the following system of equations by graphing:

\begin{equation*} \left\{ \begin{alignedat}{2} y \amp {}={} \frac{3}{2}(x-1)+4 \\ 3x-2y \amp {}={} 4 \\ \end{alignedat} \right. \end{equation*}
Explanation

The first equation, $y=\frac{3}{2}(x-1)+4\text{,}$ is a linear equation in point-slope form with a slope of $\frac{3}{2}$ that passes through the point $(1,4)\text{.}$ The second equation, $3x-2y=4\text{,}$ is a linear equation in standard form To graph this line, we either need to find the intercepts or put the equation into slope-intercept form. Just for practice, we will put the line in slope-intercept form.

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

We'll use this information to graph both lines: The two lines never intersect: they are parallel. So there are no solutions to the system of equations. We write the solution set as $\emptyset\text{.}$

### Subsection5.4.2Substitution

In Section 5.2, we covered the substitution method of solving systems of equations. We isolated one variable in one equation and then substituted into the other equation to solve for one variable.

###### Example5.4.5Solving Systems of Equations Using Substitution

Solve this system of equations using substitution:

\begin{align*} \left\{ \begin{alignedat}{4} -5x\amp {}+{} \amp 6y \amp {}={} \amp -10 \\ 4x\amp {}-{} \amp 3y \amp {}={} \amp -1 \\ \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 best to solve for $y$ in the second equation. The coefficient of $y$ in that equation is smallest.

\begin{align*} 4x-3y\amp=-1\\ -3y\amp=-1-4x\\ \divideunder{-3y}{-3}\amp=\divideunder{-1}{-3}-\divideunder{4x}{-3}\\ y\amp=\frac{1}{3}+\frac{4}{3}x \end{align*}

Replace $\substitute{y}$ in the first equation with $\substitute{\frac{1}{3}+\frac{4}{3}x}\text{,}$ giving us a linear equation in only one variable, $x\text{,}$ that we may solve:

\begin{align*} -5x+6\substitute{y}\amp=-10\\ -5x+6\left(\substitute{\frac{1}{3}+\frac{4}{3}x}\right)\amp=-10\\ -5x+2+8x\amp=-10\\ 3x+2\amp=-10\\ 3x\amp=-12\\ x\amp=-4 \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{1}{3}+\frac{4}{3}\substitute{x}\\ y\amp=\frac{1}{3}+\frac{4}{3}\left(\substitute{-4}\right)\\ y\amp=\frac{1}{3}-\frac{16}{3}\\ y\amp=-\frac{15}{3}\\ y\amp=-5 \end{align*}

To check this solution, we replace $\substitute{x}$ with $\substitute{-4}$ and $\lighthigh{y}$ with $\lighthigh{-5}$ in each equation:

\begin{align*} -5\substitute{x}+6\lighthigh{y} \amp=-10 \amp 4\substitute{x}-3\lighthigh{y} \amp= -1\\ -5(\substitute{-4})+6(\lighthigh{-5}) \amp\stackrel{?}{=}-10 \amp 4(\substitute{-4})-3(\lighthigh{-4}) \amp\stackrel{?}{=}-1\\ 20-30 \amp\stackrel{\checkmark}{=}-10 \amp -16+15 \amp\stackrel{\checkmark}{=} -1 \end{align*}

We conclude then that this system of equations is true when $x=-4$ and $y=-5\text{.}$ Our solution is the point $(-4,-5)$ and we write the solution set as $\{(-4,-5)\}\text{.}$

###### Example5.4.6Applications of Systems of Equations

The Rusk Ranch Nature Center 1 ruskranchnaturecenter.org in south-western Oregon is a volunteer run nonprofit that exists to promote the wellbeing of the local communities and conserve local nature with an emphasis on native butterflies. They sell admission tickets: $\6$ for adults and $\4$ for children. Amanda, who was working at the front desk, counted that one day she sold a total of $79$ tickets and had $\384$ in the register from those ticket sales. She didn't bother to count how many were adult tickets and how many were child tickets because she knew she could use math to figure it out at the end of the day. So, how many of the $79$ tickets were adult and how many were child?

Explanation

Let's let $a$ represent the number of adult tickets sold and $c$ represent the number of child tickets sold. We need to build two equations to solve a system for both variables.

The first equation we will build relates to the fact that there were $79$ total tickets sold. If we combine both the number of adult tickets and child tickets, the total is $79\text{.}$ This fact becomes:

\begin{equation*} a+c=79 \end{equation*}

For the second equation we need to use the per-ticket dollar amounts to generate the total cost of $\384\text{.}$ The amount of money that was made from adult tickets is found my multiplying the number of adult tickets sold, $a\text{,}$ by the price per ticket, $\6\text{.}$ Similarly, the amount of money from child tickets is found my multiplying the number of child tickets sold, $c\text{,}$ by the price per ticket, $\4\text{.}$ These two amounts will add to be $\384\text{.}$ This fact becomes:

\begin{equation*} 6a+4c=384 \end{equation*}

And so, our system is

\begin{align*} \left\{ \begin{alignedat}{4} a\amp {}+{} \amp c \amp {}={} \amp 79 \\ 6a\amp {}+{} \amp 4c \amp {}={} \amp 384 \\ \end{alignedat} \right. \end{align*}

To solve, we will use the substitution method and solve the first equation for the variable $a\text{.}$

\begin{align*} a+c\amp=79\\ a\amp=79-c \end{align*}

Now we will substitute $\substitute{79-c}$ for $\substitute{a}$ in the second equation.

\begin{align*} 6\substitute{a}+4c\amp=384\\ 6\left(\substitute{79-c}\right)+4c\amp=384\\ 474-6c+4c\amp=384\\ 474-2c\amp=384\\ -2c\amp=-90\\ c\amp=45 \end{align*}

Last, we will solve for $a$ by substituting $\substitute{45}$ in for $\substitute{c}$ in the equation $a=79-c\text{.}$

\begin{align*} a\amp=79-c\\ a\amp=79-\substitute{45}\\ a\amp=34 \end{align*}

Our conclusion is that Amanda sold $34$ adult tickets and $45$ child tickets.

###### Example5.4.7Solving Special Systems of Equations with Substitution

Solve the systems of linear equations using substitution.

1. \left\{ \begin{alignedat}{1} 3x-5y \amp {}={} 9 \\ x\amp {}={} \frac{5}{3}y+3 \\ \end{alignedat} \right.

2. \left\{ \begin{alignedat}{1} y+7\amp {}={} 4x \\ 2y-8x\amp {}={} 7 \\ \end{alignedat} \right.

Explanation

To solve the systems using substitution, we first need to solve for one variable in one equation, then substitute into the other equation.

1. In this case, $x$ is already solved for in the second equation so we can substitute $\substitute{\frac{5}{3}y+3}$ everywhere we see $\substitute{x}$ in the first equation. Then simplify and solve for $y\text{.}$

\begin{align*} 3\substitute{x}-5y\amp=9\\ 3\left(\substitute{\frac{5}{3}y+3}\right)-5y\amp=9\\ 5y+9-5y\amp=9\\ 9\amp=9 \end{align*}

We will stop here since we have eliminated all of the variables in the equation and ended with a true statement. Since $9$ always equals $9\text{,}$ no matter what, then any value of $y$ must make the original equation, $3\left(\frac{5}{3}y+3\right)-5y=9$ true. If you recall from the section on substitution, this means that both lines $3x-5y=9$ and $x=\frac{5}{3}y+3$ are in fact the same line. Since a solution to a system of linear equations is any point where the lines touch, all points along both lines are solutions. We can say this in English as, “The solutions are all points on the line $3x-5y=9\text{,}$” or in math as, “The solution set is $\{(x,y)|3x-5y=9\}\text{.}$”

2. We will first solve the top equation for $y\text{.}$

\begin{align*} y+7\amp=4x\\ y\amp=4x-7 \end{align*}

Now we can substitute $\substitute{4x-7}$ wherever we see $\substitute{y}$ in the second equation.

\begin{align*} 2\substitute{y}-8x\amp=7\\ 2\left(\substitute{4x-7}\right)-8x\amp=7\\ 8x-14-8x\amp=7\\ -14\amp=7 \end{align*}

We will stop here since we have eliminated all of the variables in the equation and ended with a false statement. Since $-14$ never equals $7\text{,}$ then no values of $x$ and $y$ can make the original system true. If you recall from the section on substitution, this means that the lines $y+7=4x$ and $2y-8x=7$ are parallel. Since a solution to a system of linear equations is any point where the lines touch, and parallel lines never touch, no points are solutions. We can say this in English as, “There are no solutions,” or in math as, “The solution set is $\emptyset\text{.}$”

### Subsection5.4.3Elimination

In Section 5.3, we explored a third way of solving systems of linear equations called elimination where we add two equations together to cancel a variable.

###### Example5.4.8Solving Systems of Equations by Elimination

Solve the system using elimination.

\begin{equation*} \left\{ \begin{alignedat}{4} 4x\amp {}-{} \amp 6y \amp {}={} \amp 13 \\ 5x\amp {}+{} \amp 4y \amp {}={} \amp -1 \\ \end{alignedat} \right. \end{equation*}
Explanation

To solve the system using elimination, we first need to scale one or both of the equations so that one variable has equal but opposite coefficients in the system. In this case, we will choose to make $y$ have opposite coefficients because the signs are already opposite for that variable in the system.

We need to multiply the first equation by $2$ and the second equation by $3\text{.}$

\begin{align*} \left\{ \begin{alignedat}{4} 4x\amp {}-{} \amp 6y \amp {}={} \amp 13 \\ 5x\amp {}+{} \amp 4y \amp {}={} \amp -1 \\ \end{alignedat} \right.\\ \left\{ \begin{alignedat}{4} \multiplyleft{2}(4x\amp {}-{} \amp 6y) \amp {}={} \amp \multiplyleft{2}(13) \\ \multiplyleft{3}(5x\amp {}+{} \amp 4y) \amp {}={} \amp \multiplyleft{3}(-1) \\ \end{alignedat} \right.\\ \left\{ \begin{alignedat}{4} 8x\amp {}-{} \amp 12y \amp {}={} \amp 26\\ 15x\amp {}+{} \amp 12y \amp {}={} \amp -3 \\ \end{alignedat} \right. \end{align*}

We now have an equivalent system of equations where the $y$-terms can be eliminated:

\begin{align*} \begin{aligned} 8x-12y\\ {}+15x+12y \end{aligned} \amp= \begin{aligned} 26\\ {}+(-3) \end{aligned}\\ \end{align*}

So we have:

\begin{align*} 23x\amp=23\\ x\amp=1 \end{align*}

To solve for $y\text{,}$ we will substitute $\substitute{1}$ for $\substitute{x}$ into either of the original equations. We will use the first equation, $4x-6y=13\text{:}$

\begin{align*} 4\substitute{x}-6y\amp=13\\ 4(\substitute{1})-6y\amp=13\\ 4-6y\amp=13\\ -6y\amp=9\\ \divideunder{-6y}{-6}\amp=\divideunder{9}{-6}\\ y\amp=-\frac{3}{2} \end{align*}

To verify this, we substitute the $x$ and $y$ values into both of the original equations.

\begin{align*} 4x-6y\amp=13\amp 5x+4y\amp=-1\\ 4(\substitute{1})-6\left(\substitute{-\frac{3}{2}}\right)\amp\stackrel{?}{=}13\amp 5(\substitute{1})+4\left(\substitute{-\frac{3}{2}}\right)\amp\stackrel{?}{=}-1\\ 4+9\amp\stackrel{\checkmark}{=}13\amp 5-6\amp\stackrel{\checkmark}{=}-1 \end{align*}

So the solution is the point $\left(-\frac{3}{2},1\right)$ and the solution set is $\left\{\left(-\frac{3}{2},1\right)\right\}\text{.}$

###### Example5.4.9Solving Special Systems of Equations with Elimination

Solve the system of equations using the elimination method.

\begin{align*} \left\{ \begin{alignedat}{4} 24x \amp {}+{} \amp 6y \amp {}={} \amp 9 \\ 8x \amp {}+{} \amp 2y \amp {}={} \amp 2 \\ \end{alignedat} \right. \end{align*}
Explanation

To eliminate the $x$-terms, we will scale the second equation by $-3\text{.}$

\begin{align*} \amp\left\{ \begin{alignedat}{4} 24x \amp {}+{} \amp 6y \amp {}={} \amp 9 \\ \multiplyleft{-3}(8x\amp {}+{} \amp 2y) \amp {}={}\amp \multiplyleft{-3}(2) \\ \end{alignedat} \right.\\ \amp\left\{ \begin{alignedat}{4} 24x \amp {}+{} \amp 6y \amp {}={} \amp 9 \\ -24x \amp {}-{} \amp 6y \amp {}={} \amp -6 \\ \end{alignedat} \right. \end{align*}

Adding the respective sides of the equation, we have:

\begin{equation*} 0=3 \end{equation*}

Both of the variables have been eliminated. In this case, the statement $0=3$ is just false, no matter what $x$ and $y$ are. So the system has no solution. The solution set is $\emptyset\text{.}$

###### Example5.4.10Deciding to Use Substitution versus Elimination

Decide which method would be easiest to solve the systems of linear equations: substitution or elimination.

1. \left\{ \begin{alignedat}{4} 2x \amp {}+{} \amp 3y \amp {}={} \amp -11 \\ 5x \amp {}-{} \amp 6y \amp {}={} \amp 13 \\ \end{alignedat} \right.

2. \left\{ \begin{alignedat}{4} x \amp {}-{} \amp 7y \amp {}={} \amp 10 \\ 9x \amp {}-{} \amp 16y \amp {}={} \amp -4 \\ \end{alignedat} \right.

3. \left\{ \begin{alignedat}{4} 6x \amp {}+{} \amp 30y \amp {}={} \amp 15 \\ 4x \amp {}+{} \amp 20y \amp {}={} \amp 10 \\ \end{alignedat} \right.

4. \left\{ \begin{alignedat}{1} y\amp {}={} 3x {}-{} 2 \\ y\amp {}={} 7x {}+{} 6 \\ \end{alignedat} \right.

Explanation
1. Elimination is probably easiest here. Multiply the first equation by $2$ and eliminate the $y$ variables. The solution to this one is $(-1,-3)$ if you want to solve it for practice.

2. Substitution is probably easiest here. Solve the first equation for $x$ and substitute it into the second equation. We could use elimination if we multiplied the first equation by $-9$ and eliminate the $x$ variable, but it's probably a little more work than substitution. The solution to this one is $(-4,-2)$ if you want to solve it for practice.

3. Elimination is probably easiest here. Multiply the first equation by $2$ and the second equation by $-3\text{.}$ Doing this will eliminate both variables and leave you with $0=0\text{.}$ This should mean that all points on the line are solutions. So the solution set is $\{(x,y)|6x+30y=15\}\text{.}$

4. Substitution is definitely easiest here. Substituting $y$ from one equation into $y$ in the other equation gives you $3x-2=7x+6\text{.}$ Solve this and find then find $y$ and you should get the solution to the system to be $(-2,-8)$ if you want to solve it for practice.

### Subsection5.4.4Exercises

###### Solving Systems of Linear Equations by Graphing

Use a graph to solve the system of equations.

###### 1

\left\{\begin{aligned} x+y\amp=0\\ 3x-y\amp=8 \end{aligned}\right.

###### 2

\left\{\begin{aligned} 4x-2y\amp=4\\ x+2y\amp=6 \end{aligned}\right.

###### 3

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

###### 4

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

###### 5

\left\{\begin{aligned} -10x+15y\amp=60\\ 6x-9y\amp=36 \end{aligned}\right.

###### 6

\left\{\begin{aligned} 6x-8y\amp=32\\ 9x-12y\amp=12 \end{aligned}\right.

###### 7

\left\{\begin{aligned} y\amp=-\frac{3}{5}x+7\\ 9x+15y\amp=105 \end{aligned}\right.

###### 8

\left\{\begin{aligned} 9y-12x\amp=18\\ y\amp=\frac{4}{3}x+2 \end{aligned}\right.

###### 9

Solve the following system of equations.

\begin{equation*} \left\{ \begin{aligned} {c} \amp = {-46 - 5C}\\ {2C+5c} \amp = {0} \end{aligned} \right. \end{equation*}
###### 10

Solve the following system of equations.

\begin{equation*} \left\{ \begin{aligned} {p} \amp = {-2a}\\ {4a+2p} \amp = {0} \end{aligned} \right. \end{equation*}
###### 11

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {2x+2y} \amp = {18} \\ {4x+3y} \amp = {26} \end{aligned}\right. \end{equation*}
###### 12

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {x+3y} \amp = {8} \\ {3x+2y} \amp = {10} \end{aligned}\right. \end{equation*}
###### 13

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {2x+5y} \amp = {-17} \\ {2x-4y} \amp = {28} \end{aligned}\right. \end{equation*}
###### 14

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {4x-4y} \amp = {-12} \\ {3x+3y} \amp = {45} \end{aligned}\right. \end{equation*}
###### 15

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {5x+3y} \amp = 2 \\ {-20x-12y} \amp = -8 \end{aligned}\right. \end{equation*}
###### 16

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {x+2y} \amp = 2 \\ {3x+6y} \amp = 6 \end{aligned}\right. \end{equation*}
###### 17

A rectangle’s length is $6$ feet longer than five times its width. The rectangle’s perimeter is $192$ feet. Find the rectangle’s length and width.

The rectangle’s length is feet, and its width is feet.

###### 18

A school fund raising event sold a total of $178$ tickets and generated a total revenue of ${\420.90}\text{.}$ There are two types of tickets: adult tickets and child tickets. Each adult ticket costs ${\4.45}\text{,}$ and each child ticket costs ${\1.55}\text{.}$ Write and solve a system of equations to answer the following questions.

adult tickets and child tickets were sold.

###### 19

A test has $21$ problems, which are worth a total of $78$ points. There are two types of problems in the test. Each multiple-choice problem is worth $3$ points, and each short-answer problem is worth $6$ points. Write and solve a system equation to answer the following questions.

This test has multiple-choice problems and short-answer problems.

###### 20

A test has $20$ problems, which are worth a total of $130$ points. There are two types of problems in the test. Each multiple-choice problem is worth $3$ points, and each short-answer problem is worth $10$ points. Write and solve a system equation to answer the following questions.

This test has multiple-choice problems and short-answer problems.

###### 21

Kristen invested a total of ${\8{,}000}$ in two accounts. One account pays $5\%$ interest annually; the other pays $4\%$ interest annually. At the end of the year, Kristen earned a total of ${\355}$ in interest. Write and solve a system of equations to find how much money Kristen invested in each account.

Kristen invested in the $5\%$ account and in the $4\%$ account.

###### 22

Lily invested a total of ${\13{,}000}$ in two accounts. After a year, one account lost $6.8\%\text{,}$ while the other account gained $2.5\%\text{.}$ In total, Lily lost ${\419}\text{.}$ Write and solve a system of equations to find how much money Lily invested in each account.

Lily invested in the account with $6.8\%$ loss and in the account with $2.5\%$ gain.

###### 23

Town A and Town B were located close to each other, and recently merged into one city. Town A had a population with $12\%$ African Americans. Town B had a population with $6\%$ African Americans. After the merge, the new city has a total of $4000$ residents, with $8.1\%$ African Americans. Write and solve a system of equations to find how many residents Town A and Town B used to have.

Town A used to have residents, and Town B used to have residents.

###### 24

You poured some $12\%$ alcohol solution and some $6\%$ alcohol solution into a mixing container. Now you have $600$ grams of $8 \%$ alcohol solution. How many grams of $12\%$ solution and how many grams of $6 \%$ solution did you pour into the mixing container?

Write and solve a system equation to answer the following questions.

You mixed grams of $12\%$ solution with grams of $6\%$ solution.

###### 25

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {x+3y} \amp = {-37} \\ {3x+3y} \amp = {-57} \end{aligned}\right. \end{equation*}
###### 26

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {3x+2y} \amp = {-14} \\ {3x+5y} \amp = {1} \end{aligned}\right. \end{equation*}
###### 27

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {6x-3y} \amp = {-24} \\ {3x+5y} \amp = {-25} \end{aligned}\right. \end{equation*}
###### 28

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {3x+4y} \amp = {-45} \\ {-3x+3y} \amp = {-18} \end{aligned}\right. \end{equation*}
###### 29

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {3x+4y} \amp = -2 \\ {-12x-16y} \amp = -2 \end{aligned}\right. \end{equation*}
###### 30

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {3x+2y} \amp = -2 \\ {-6x-4y} \amp = -2 \end{aligned}\right. \end{equation*}
###### 31

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {4x+y} \amp = -2 \\ {12x+3y} \amp = -6 \end{aligned}\right. \end{equation*}
###### 32

Solve the following system of equations.

\begin{equation*} \left\{\begin{aligned} {4x+4y} \amp = -3 \\ {8x+8y} \amp = -6 \end{aligned}\right. \end{equation*}
###### 33

A test has $20$ problems, which are worth a total of $112$ points. There are two types of problems in the test. Each multiple-choice problem is worth $5$ points, and each short-answer problem is worth $7$ points. Write and solve a system of equations to answer the following questions.

This test has multiple-choice problems and short-answer problems.

###### 34

Wenwu invested a total of ${\5{,}000}$ in two accounts. One account pays $7\%$ interest annually; the other pays $6\%$ interest annually. At the end of the year, Wenwu earned a total of ${\345}$ in interest. Write and solve a system of equations to find how much money Wenwu invested in each account.

Wenwu invested in the $7\%$ account and in the $6\%$ account.

###### 35

Town A and Town B were located close to each other, and recently merged into one city. Town A had a population with $6\%$ Hispanics. Town B had a population with $8\%$ Hispanics. After the merge, the new city has a total of $5000$ residents, with $7.52\%$ Hispanics. Write and solve a system of equations to find how many residents Town A and Town B used to have.

Town A used to have residents, and Town B used to have residents.

###### 36

You poured some $6\%$ alcohol solution and some $10\%$ alcohol solution into a mixing container. Now you have $800$ grams of $8.4 \%$ alcohol solution. Write and solve a system of equations to find how many grams of $6\%$ solution and how many grams of $10 \%$ solution you poured into the mixing container.

You mixed grams of $6\%$ solution with grams of $10\%$ solution.

###### 37

If a boat travels from Town A to Town B, it has to travel ${797.5\ {\rm mi}}$ along a river. A boat traveled from Town A to Town B along the river’s current with its engine running at full speed. This trip took ${27.5\ {\rm hr}}\text{.}$ Then the boat traveled back from Town B to Town A, again with the engine at full speed, but this time against the river’s current. This trip took ${72.5\ {\rm hr}}\text{.}$ Write and solve a system of equations to answer the following questions.

The boat’s speed in still water with the engine running at full speed is .

The river current’s speed was .

###### 38

A small fair charges different admission for adults and children. It charges ${\3.50}$ for adults, and ${\2}$ for children. On a certain day, the total revenue is ${\5{,}957.50}$ and the fair admits ${2300}$ people. How many adults and children were admitted?

There were adults and children at the fair.