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4.4: Applications with systems of equations

Last updated

Oct 3, 2021
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- 4.3: System of Equations - The Addition Method
- 4.5: Systems of three linear equations in three variables

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We saw these types of examples in a previous chapter, but with one variable. In this section, we review the same types of applications, but solving in a more sophisticated way using systems of equations. Once we set up the system, we can solve using any method we choose. However, setting up the system may be the challenge, but as long as we follow the method we used before, we will be fine. We use tables to organize the parameters.

Value & Interest Problems

Example 4.4.1

There were $41$ tickets sold for an event. Tickets for children cost $$1.50$ and tickets for adults cost $$2.00$ . Total receipts for the event were $$73.50$ . How many of each type of ticket were sold?

Solution

First, we can make a table to organize the given information and then create an equation. Let $c$ represent the number of children tickets sold and $a$ represent the number of adult tickets sold.

Table 4.4.1
	Amount	Value (in $)	Total Value
Adult tickets	$a$	$$2$	$2a$
Children tickets	$c$	$$1.50$	$1.5c$
Total	$41$		$$73.50$

Now let’s set up the system. The total number of tickets sold was $41$ and the total revenue from the tickets was $$73.50$ , and so we obtain the system $\begin{aligned}a+c&=41 \\ 2a+1.5c&=73.50\end{aligned}$

At this point, we can solve using any method we choose. Since the coefficient of $a$ and $c$ in the first equation are both one, then let’s use the method of substitution. We will solve for $a$ in the first equation:

$\begin{aligned}a+c&=41 \\ a&=41-c\end{aligned}$

Now, we can substitute $a$ into the second equation and solve:

$\begin{array}{rl}2a+1.5c=73.50&\text{Plug-n-chug }a=41-c \\ 2\color{blue}{(41-c)}\color{black}{}+1.5c=73.50&\text{Distribute} \\ 82-2c+1.5c=73.50&\text{Combine like terms} \\ 82-0.5c=73.50&\text{Isolate the variable term} \\ -0.5c=-8.50&\text{Multiply by the reciprocal of }-0.5 \\ c=17&\text{Number of children tickets}\end{array}\nonumber$

Since $c = 17$ , then we can plug-n-chug $c = 17$ into one of the equations to obtain $a$ :

$\begin{array}{rl}a=41-c&\text{Plug-n-chug }c=17 \\ a=41-\color{blue}{(17)}\color{black}{}&\text{Evaluate} \\ a=24&\text{Number of adult tickets}\end{array}\nonumber$

Thus, there were $17$ children tickets and $24$ adult tickets sold.

Example 4.4.2

Aaron invests $$9,700$ in two different accounts. The first account paid $7$ %, the second account paid $11$ % in interest. At the end of the first year he had earned $$863$ in interest. How much was in each account?

Solution

First, we can make a table to organize the given information and then create an equation. Let $x$ represent the amount of investment in the first account and $y$ represent the amount of investment in the second account.

Table 4.4.2
	Principal	rate	Total Interest
Account 1	$x$	$0.07$	$0.07x$
Account 2	$y$	$0.11$	$0.11y$
Total	$$9,700$		$$863$

Now let’s set up the system. The total interest made in the one year was $863 and the total invested was $9,700, so we obtain the system $\begin{aligned}x+y&=9700 \\ 0.07x+0.11y&=863\end{aligned}$

At this point, we can solve using any method we choose. Since the coefficient of $x$ and $y$ in the first equation are both one, then let’s use the method of substitution. We will solve for $y$ in the first equation: $\begin{aligned}x+y&=9700 \\ y&=9700-x\end{aligned}$

Now, we can substitute $y$ into the second equation and solve:

$\begin{array}{rl}0.07x+0.11\color{blue}{(9700-x)}\color{black}{}=863&\text{Distribute} \\ 0.07x+1067-0.11x=863&\text{Combine like terms} \\ -0.04x+1067=863&\text{Isolate the variable term} \\ -0.04x=-204&\text{Multiply by the reciprocal of }-0.04 \\ x=5100&\text{Investment ammount for Account 1}\end{array}\nonumber$

Since the investment amount for Account 1 was $$5,100$ , then the investment amount for Account 2 was $$4,600$ $(9700 − 5100 = 4600)$ . Thus, the investment amounts for Account 1 and Account 2 was $$5,100$ and $$4,600$ , respectively.

Mixture Problems

Example 4.4.3

A farmer has two types of milk, one that is $24$ % butterfat and another which is $18$ % butterfat. How much of each should he use to end up with $42$ gallons of $20$ % butterfat?

Solution

First, we can make a table to organize the given information and then create a system. Let $x$ represent the number of gallons of the $24$ % butterfat milk and $y$ represent the number of gallons of the $18$ % butterfat milk.

Table 4.4.3
	Amount	Concentration	Total Butterfat
24% butterfat	$x$	$0.24$	$0.24x$
18% butterfat	$y$	$0.18$	$0.18y$
20% butterfat	$42$	$0.20$	$0.20(42)$

Now let’s set up the system:

$\begin{aligned}x+y&=42 \\ 0.24x+0.18y&=8.4\end{aligned}$

At this point, we can solve using any method we choose. Let’s solve using elimination. We can choose to eliminate $x$ and we will multiply the first equation by $−0.24$ :

$\begin{array}{rl}\color{blue}{-0.24}\color{black}{}\cdot (x+y)=(42)\cdot\color{blue}{-0.24}\color{black}{}&\text{Distribute} \\ -0.24x-0.24y=-10.08\end{array}\nonumber$

Notice the $x$ variable terms have the same coefficients with opposite signs. Now we can add and eliminate $x$ :

$\begin{array}{l} -0.24x-0.24y=-10.08 \\ \underline{+\quad 0.24x+0.18y=8.4}\:\: \\ \qquad\qquad -0.06y=-1.68 \end{array}\qquad \text{ Add the equations}\nonumber$

Now, we can easily solve as usual:

$\begin{array}{rl}-0.06y=-1.68&\text{Multiply by the reciprocal of }-0.06 \\ y=28&\text{Number of gallons from the }18\text{% butterfat} \end{array}\nonumber$

Since the number of gallons from the $18$ % butterfat milk was $28$ , then the number of gallons from the $24$ % butterfat milk was $14 (42 − 28 = 14)$ . Thus, the farmer will need $14$ gallons of the $24$ % butterfat milk and $28$ gallons of the $18$ % butterfat milk to make $42$ gallons of a $20$ % butterfat milk.

Example 4.4.4

A solution of pure antifreeze is mixed with water to make a $65$ % antifreeze solution. How much of each should be used to make $70$ liters?

Solution

Mixture problems with a pure solution or water contains no other chemicals. For pure solutions, the percentage is $100$ % (or $1$ in the table) and for water, the percentage is $0$ %. First, we can make a table to organize the given information and then create a system. Let $a$ represent the number of liters of antifreeze and $w$ represent the number of liters of water.

Table 4.4.4
	Amount	Concentration	Total Butterfat
Antifreeze	$a$	$1$	$1a$
Water	$w$	$0$	$0$
65% Solution	$70$	$0.65$	$0.65(70)$

Now let’s set up the system:

$\begin{aligned}a+w&=70 \\ 1a&=45.5\end{aligned}$

At this point, we can solve using any method we choose. Since we see from the system that $a = 45.5$ , then let’s solve by substitution. We can put the second equation in for $a$ in the first equation:

$\begin{array}{rl}a+w=70&\text{Plug-n-chug }a=45.5 \\ \color{blue}{45.5}\color{black}{}+w=70&\text{Isolate the variable term} \\ w=24.5&\text{The number of liters in water}\end{array}\nonumber$

Thus, the number of liters of water needed is $24.5$ liters and antifreeze needed is $45.5$ liters.

Example 4.4.5

In a candy shop, chocolate, which sells for $$4$ per pound, is mixed with nuts, which are sold for $$2.50$ per pound, to form a chocolate-nut candy which sells for $$3.50$ a pound. How many pounds of each are used to make $30$ pounds of the mixture?

Solution

First, we can make a table to organize the given information and then create an equation. Let $c$ represent the number of pounds of chocolate and $n$ represent the number of pounds of nuts.

Table 4.4.5
	Amount	Cost	Total Cost
Chocolate	$c$	$$4$	$4c$
Nuts	$n$	$$2.50$	$2.5n$
Mix	$30$	$$3.50$	$3.5(30)$

Now let’s set up the system:

$\begin{aligned}c+n&=30 \\ 4c+2.5n&=105\end{aligned}$

At this point, we can solve using any method we choose. Let’s solve using elimination. We can choose to eliminate $c$ and we will multiply the first equation by $−4$ :

$\begin{array}{rl}\color{blue}{-4}\color{black}{}\cdot (c+n)=(3)\cdot\color{blue}{-4}\color{black}{}&\text{Distribute} \\ -4c-4n=-120\end{array}\nonumber$

Notice the $c$ variable terms have the same coefficients with opposite signs. Now we can add and eliminate $c$ :

$\begin{array}{l} \quad -4c-4n=-120 \\ \underline{+\quad 4c+2.5n=105}\:\: \\ \qquad\quad -1.5n=-15 \end{array}\qquad \text{ Add the equations}\nonumber$

Now, we can easily solve as usual:

$\begin{array}{rl}-1.5n=-15&\text{Multiply by the reciprocal of }-1.5 \\ n=10&\text{Number of pounds of nuts}\end{array}\nonumber$

Since we need $10$ pounds of nuts, then this implies that we need $20$ pounds of chocolate $(30−10 = 20)$ .

Uniform Motion with Unknown Rates

When we looked at uniform motion in a prior chapter, we always were given something about the rate. However, now we discuss uniform motion with a wind force and a water current rates, where we know very little and will have to use a system to solve for the rates.

Example 4.4.6

Turkey the Pigeon travels the same distance of $72$ miles in $4$ hours against the wind as it does traveling $3$ hours with the wind in local skies. What is the rate of Turkey the Pigeon in still air and the rate of the wind?

Solution

First, we can make a table to organize the given information and then create an equation. Let $r$ represent the rate of Turkey in still air and $w$ represent the rate of the wind. If Turkey travels with the wind, then Turkey is getting a little push from the wind, meaning traveling a little faster. If Turkey travels against the wind, then Turkey is getting a little push back from the wind, meaning traveling a little slower.

Table 4.4.6
	rate	time	distance
with the wind	$r+w$	$3$	$3(r+w)$
against the wind	$r-w$	$4$	$4(r-w)$

Since Turkey is traveling a distance of $72$ miles, then this is the distance for both routes. Now let’s set up the system:

$\begin{aligned}3(r+w)&=72 \\ 4(r-w)&=72\end{aligned}$

Since $3$ is a factor of $72$ and $4$ is a factor of $72$ , let’s divide each side of each equation:

$\begin{array}{r} \underline{3(r+w)=72} \\ 3\qquad 3 \\ r+w =24\end{array}\nonumber$

$\begin{array}{r} \underline{4(r-w)=72} \\ 4\qquad 4 \\ r-w =18\end{array}\nonumber$

Notice the $w$ variable terms have the same coefficients with opposite signs. Now we can add and eliminate $w$ :

$\begin{array}{l} \qquad r+w=24 \\ \underline{+\quad r-w=18}\:\: \\ \qquad\quad 2r=42 \end{array}\qquad \text{ Add the equations}\nonumber$

Now, we can easily solve as usual:

$\begin{array}{rl}2r=42 &\text{Multiply by the reciprocal of }2 \\ r=21&\text{Rate of Turkey in still air}\end{array}\nonumber$

Since Turkey’s rate is $21$ miles per hour in still air, then this implies that the rate of the wind is $3$ miles per hour $(24 − 21 = 3)$ .

Example 4.4.7

A boat travels upstream for $156$ miles in $3$ hours and returns in $2$ hours traveling downstream in a local stream of water. What is the rate of the boat in still water and the rate of the current?

Solution

First, we can make a table to organize the given information and then create an equation. Let $r$ represent the rate of boat in still water and $c$ represent the rate of the current. If the boat travels with the current, then the boat is getting a little push from the current, meaning traveling a little faster. If the boat travels against the current, then the boat is getting a little push back from the current, meaning traveling a little slower.

Table 4.4.7
	rate	time	distance
with the current	$r+c$	$2$	$2(r+c)$
against the current	$r-c$	$3$	$3(r-c)$

Since the boat is traveled a distance of $156$ miles, then this is the distance for both routes. Now let’s set up the system:

$\begin{aligned}2(r+c)&=156 \\ 3(r-c)&=156\end{aligned}$

Since $2$ is a factor of $156$ and $3$ is a factor of $156$ , let’s divide each side of each equation:

$\begin{array}{r} \underline{2(r+c)=156} \\ 2\qquad 2 \\ r+c =78\end{array}\nonumber$

$\begin{array}{r} \underline{3(r-c)=156} \\ 3\qquad 3 \\ r-c =32\end{array}\nonumber$

Notice the $c$ variable terms have the same coefficients with opposite signs. Now we can add and eliminate $c$ :

$\begin{array}{l} \qquad r+c=78 \\ \underline{+\:\: \quad r-c=52}\:\: \\ \qquad\quad 2r=130 \end{array}\qquad \text{ Add the equations}\nonumber$

Now, we can easily solve as usual:

$\begin{array}{rl}2r=130&\text{Multiply by the reciprocal of }2 \\ r=65&\text{Rate of the boat in still water}\end{array}\nonumber$

Since the rate of the boat in still water is $65$ miles per hour, then this implies that the rate of the current is $13$ miles per hour $(65 − 52 = 13)$ .

Applications with Systems of Equations Homework

Exercise 4.4.1

The attendance at a school concert was $578$ . Admission was $$2.00$ for adults and $$1.50$ for children. The total receipts were $$985.00$ . How many adults and how many children attended?

Exercise 4.4.2

There were $429$ people at a play. Admission was $$1$ each for adults and $75$ cents each for children. The receipts were $$372.50$ . How many children and how many adults attended?

Exercise 4.4.3

There were $200$ tickets sold for a women’s basketball game. Tickets forstudents were $50$ cents each and for adults $75$ cents each. The total amount of money collected was $$132.50$ . How many of each type of ticket was sold?

Exercise 4.4.4

There were $203$ tickets sold for a volleyball game. For activity-card holders, the price was $$1.25$ each and for non-card holders the price was $$2$ each. The total amount of money collected was $$310$ . How many of each type of ticket was sold?

Exercise 4.4.5

At a local ball game the hot dogs sold for $$2.50$ each and the hamburgers sold for $$2.75$ each. There were $131$ total hamburgers and hot dogs sold for a total value of $$342$ . How many of each was sold?

Exercise 4.4.6

At a recent Vikings game, there was $$445$ in admission tickets. The cost of a student ticket was $$1.50$ and the cost of a non-student ticket was $$2.50$ . A total of $232$ tickets were sold. How many students and how many non-students attended the game?

Exercise 4.4.7

A total of $$27,000$ is invested, part of it at $12$ % and the rest at $13$ %. The total interest after one year is $$3,385$ . How much was invested at each rate?

Exercise 4.4.8

A total of $$50,000$ is invested, part of it at $5$ % and the rest at $7.5$ %. The total interest after one year is $$3,250$ . How much was invested at each rate?

Exercise 4.4.9

A total of $$9,000$ is invested, part of it at $10$ % and the rest at $12$ %. The total interest after one year is $$1,030$ . How much was invested at each rate?

Exercise 4.4.10

A total of $$18,000$ is invested, part of it at $6$ % and the rest at $9$ %. The total interest after one year is $$1,248$ . How much was invested at each rate?

Exercise 4.4.11

An inheritance of $$10,000$ is invested in $2$ ways, part at $9.5%$ and the remainder at $11$ %. The combined annual interest was $$1,038.50$ . How much was invested at each rate?

Exercise 4.4.12

Kerry earned a total of $$900$ last year on his investments. If $$7,000$ was invested at a certain rate of return and $$9,000$ was invested in a fund with a rate that was $2$ % higher, find the two rates of interest.

Exercise 4.4.13

Jason earned $$256$ interest last year on his investments. If $$1,600$ was invested at a certain rate of return and $$2,400$ was invested in a fund with a rate that was double the rate of the first fund, find the two rates of interest.

Exercise 4.4.14

Millicent earned $$435$ last year in interest. If $$3,000$ was invested at a certain rate of return and $$4,500$ was invested in a fund with a rate that was $2$ % lower, find the two rates of interest.

Exercise 4.4.15

A total of $$8,500$ is invested, part of it at $6$ % and the rest at $3.5$ %. The total interest after one year is $$385$ . How much was invested at each rate?

Exercise 4.4.16

A total of $$12,000$ was invested, part of it at $9$ % and the rest at $7.5$ %. The total interest after one year is $$1,005$ . How much was invested at each rate?

Exercise 4.4.17

A total of $$15,000$ is invested, part of it at $8$ % and the rest at $11$ %. The total interest after one year is $$1,455$ . How much was invested at each rate?

Exercise 4.4.18

A total of $$17,500$ is invested, part of it at $7.25$ % and the rest at $6.5$ %. The total interest after one year is $$1,227.50$ . How much was invested at each rate?

Exercise 4.4.19

A total of $$6,000$ is invested, part of it at $4.25$ % and the rest at $5.75$ %. The total interest after one year is $$300$ . How much was invested at each rate?

Exercise 4.4.20

A total of $$14,000$ is invested, part of it at $5.5$ % and the rest at $9$ %. The total interest after one year is $$910$ . How much was invested at each rate?

Exercise 4.4.21

A total of $$11,000$ is invested, part of it at $6.8$ % and the rest at $8.2$ %. The total interest after one year is $$797$ . How much was invested at each rate?

Exercise 4.4.22

An investment portfolio earned $$2,010$ in interest last year. If $$3,000$ was invested at a certain rate of return and $$24,000$ was invested in a fund with a rate that was $4$ % lower, find the two rates of interest.

Exercise 4.4.23

Samantha earned $$1,480$ in interest last year on her investments. If $$5,000$ was invested at a certain rate of return and $$11,000$ was invested in a fund with a rate that was two-thirds the rate of the first fund, find the two rates of interest.

Exercise 4.4.24

Solution A is $50$ % acid and solution B is $80$ % acid. How much of each should be used to make $100$ cc. of a solution that is $68$ % acid?

Exercise 4.4.25

A certain grade of milk contains $10$ % butterfat and a certain grade of cream $60$ % butterfat. How many quarts of each must be taken so as to obtain a mixture of $100$ quarts that will be $45$ % butterfat?

Exercise 4.4.26

A farmer has some cream which is $21$ % butterfat and some which is $15$ % butterfat. How many gallons of each must be mixed to produce $60$ gallons of cream which is $19$ % butterfat?

Exercise 4.4.27

A syrup manufacturer has some pure maple syrup and some which is $85$ % maple syrup. How many liters of each should be mixed to make $150$ liters which is $96$ % maple syrup?

Exercise 4.4.28

A chemist wants to make $50$ mL of a $16$ % acid solution by mixing a $13$ % acid solution and an $18$ % acid solution. How many milliliters of each solution should the chemist use?

Exercise 4.4.29

A hair dye is made by blending $7$ % hydrogen peroxide solution and a $4$ % hydrogen peroxide solution. How many milliliters of each are used to make a $300$ mL solution that is $5$ % hydrogen peroxide?

Exercise 4.4.30

A paint that contains $21$ % green dye is mixed with a paint that contains $15$ % green dye. How many gallons of each must be used to make $60$ gallons of paint that is $19$ % green dye?

Exercise 4.4.31

A candy mix sells for $$2.20$ per kilogram. It contains chocolates worth $$1.80$ per kilogram and other candy worth $$3.00$ per kilogram. How much of each are in $15$ kilograms of the mixture?

Exercise 4.4.32

To make a weed and feed mixture, the Green Thumb Garden Shop mixes fertilizer worth $$4$ per pound with a weed killer worth $$8$ per pound. The mixture will cost $$6.00$ per pound. How much of each should be used to prepare $500$ pounds of the mixture?

Exercise 4.4.33

A grocer is mixing a $40$ -cent per pound coffee with a $60$ -cent per pound coffee to make a mixture worth $54$ ¢ per pound How much of each kind of coffee should be used to make $70$ pounds of the mixture?

Exercise 4.4.34

A grocer wishes to mix sugar at $9$ cents per pound with sugar at $6$ cents per pound to make $60$ pounds at $7$ cents per pound. What quantity of each must he take?

Exercise 4.4.35

A high-protein diet supplement that costs $$6.75$ per pound is mixed with a vitamin supplement that costs $$3.25$ per pound. How many pounds of each should be used to make $5$ pounds of a mixture that costs $$4.65$ per pound?

Exercise 4.4.36

A goldsmith combined an alloy that costs $$4.30$ per ounce with an alloy that costs $$1.80$ per ounce. How many ounces of each were used to make a mixture of $200$ ounces costing $$2.50$ per ounce?

Exercise 4.4.37

A grocery store offers a cheese and fruit sampler that combines cheddar cheese that costs $$8$ per kilogram with kiwis that cost $$3$ per kilogram. How many kilograms of each were used to make a $5$ kilogram mixture that costs $$4.50$ per kilogram?

Exercise 4.4.38

A caterer made an ice cream punch by combining fruit juice that cost $$2.25$ per gallon with ice cream that costs $$3.25$ per gallon. How many gallons of each were used to make $100$ gallons of punch costing $$2.50$ per pound?

Exercise 4.4.39

A clothing manufacturer has some pure silk thread and some thread that is $85$ % silk. How many kilograms of each must be woven together to make $75$ kilograms of cloth that is $96$ % silk?

Exercise 4.4.40

A carpet manufacturer blends two fibers, one $20$ % wool and the second $50$ % wool. How many pounds of each fiber should be woven together to produce $600$ pounds of a fabric that is $28$ % wool?

Exercise 4.4.41

The manager of a specialty food store combined almonds that cost $$4.50$ per pound with walnuts that cost $$2.50$ per pound. How many pounds of each were used to make a $100$ -pound mixture that cost $$3.24$ per pound?

Exercise 4.4.42

A tea that is $20$ % jasmine is blended with a tea that is $15$ % jasmine. How many pounds of each tea are used to make $5$ pounds of tea that is $18$ % jasmine?

Exercise 4.4.43

How many milliliters of pure chocolate must be added to $150$ mL of chocolate topping that is $50$ % chocolate to make a topping that is $75$ % chocolate?

Exercise 4.4.44

How many ounces of pure bran flakes must be added to $50$ ounces of cereal that is $40$ % bran flakes to produce a mixture that is $50$ % bran flakes?

Exercise 4.4.45

A ground meat mixture is formed by combining meat that costs $$2.20$ per pound with meat that costs $$4.20$ per pound. How many pounds of each were used to make a $50$ -pound mixture that costs $$3$ per pound?

Exercise 4.4.46

How many grams of pure water must be added to $50$ grams of pure acid to make a solution that is $40$ % acid?

Exercise 4.4.47

A lumber company combined oak wood chips that cost $$3.10$ per pound with pine wood chips that cost $$2.50$ per pound. How many pounds of each were used to make an $80$ -pound mixture costing $$2.65$ per pound?

Exercise 4.4.48

How many ounces of pure water must be added to $50$ ounces of a $15$ % saline solution to make a saline solution that is $10$ % salt?

Exercise 4.4.49

A boat travels upstream for $216$ miles in $4$ hours and returns in $3$ hours traveling downstream in a local stream of water. What is the rate of the boat in still water and the rate of the current?

Exercise 4.4.50

A boat travels upstream for $12$ miles in $3$ hours and returns in $2$ hours traveling downstream in a local stream of water. What is the rate of the boat in still water and the rate of the current?

Exercise 4.4.51

A boat travels upstream for $336$ miles in $4$ hours and returns in $3$ hours traveling downstream in a local stream of water. What is the rate of the boat in still water and the rate of the current?

Exercise 4.4.52

Turkey the Pigeon travels the same distance of $280$ miles in $5$ hours against the wind as it does traveling $4$ hours with the wind in local skies. What is the rate of Turkey the Pigeon in still air and the rate of the wind?

Exercise 4.4.53

Turkey the Pigeon travels the same distance of $24$ miles in $4$ hours against the wind as it does traveling $3$ hours with the wind in local skies. What is the rate of Turkey the Pigeon in still air and the rate of the wind?

Exercise 4.4.54

Turkey the Pigeon travels the same distance of $120$ miles in $4$ hours against the wind as it does traveling $3$ hours with the wind in local skies. What is the rate of Turkey the Pigeon in still air and the rate of the wind?

Value & Interest Problems

Mixture Problems

Uniform Motion with Unknown Rates

Applications with Systems of Equations Homework

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