4.4: Applications with systems of equations
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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
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.
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 a+c=412a+1.5c=73.50
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:
a+c=41a=41−c
Now, we can substitute a into the second equation and solve:
2a+1.5c=73.50Plug-n-chug a=41−c2(41−c)+1.5c=73.50Distribute82−2c+1.5c=73.50Combine like terms82−0.5c=73.50Isolate the variable term−0.5c=−8.50Multiply by the reciprocal of −0.5c=17Number of children tickets
Since c=17, then we can plug-n-chug c=17 into one of the equations to obtain a:
a=41−cPlug-n-chug c=17a=41−(17)Evaluatea=24Number of adult tickets
Thus, there were 17 children tickets and 24 adult tickets sold.
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.
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 x+y=97000.07x+0.11y=863
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: x+y=9700y=9700−x
Now, we can substitute y into the second equation and solve:
0.07x+0.11(9700−x)=863Distribute0.07x+1067−0.11x=863Combine like terms−0.04x+1067=863Isolate the variable term−0.04x=−204Multiply by the reciprocal of −0.04x=5100Investment ammount for Account 1
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
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.
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:
x+y=420.24x+0.18y=8.4
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:
−0.24⋅(x+y)=(42)⋅−0.24Distribute−0.24x−0.24y=−10.08
Notice the x variable terms have the same coefficients with opposite signs. Now we can add and eliminate x:
−0.24x−0.24y=−10.08+0.24x+0.18y=8.4_−0.06y=−1.68 Add the equations
Now, we can easily solve as usual:
−0.06y=−1.68Multiply by the reciprocal of −0.06y=28Number of gallons from the 18% butterfat
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.
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.
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:
a+w=701a=45.5
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:
a+w=70Plug-n-chug a=45.545.5+w=70Isolate the variable termw=24.5The number of liters in water
Thus, the number of liters of water needed is 24.5 liters and antifreeze needed is 45.5 liters.
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.
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:
c+n=304c+2.5n=105
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:
−4⋅(c+n)=(3)⋅−4Distribute−4c−4n=−120
Notice the c variable terms have the same coefficients with opposite signs. Now we can add and eliminate c:
−4c−4n=−120+4c+2.5n=105_−1.5n=−15 Add the equations
Now, we can easily solve as usual:
−1.5n=−15Multiply by the reciprocal of −1.5n=10Number of pounds of nuts
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.
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.
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:
3(r+w)=724(r−w)=72
Since 3 is a factor of 72 and 4 is a factor of 72, let’s divide each side of each equation:
3(r+w)=72_33r+w=24
4(r−w)=72_44r−w=18
Notice the w variable terms have the same coefficients with opposite signs. Now we can add and eliminate w:
r+w=24+r−w=18_2r=42 Add the equations
Now, we can easily solve as usual:
2r=42Multiply by the reciprocal of 2r=21Rate of Turkey in still air
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).
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.
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:
2(r+c)=1563(r−c)=156
Since 2 is a factor of 156 and 3 is a factor of 156, let’s divide each side of each equation:
2(r+c)=156_22r+c=78
3(r−c)=156_33r−c=32
Notice the c variable terms have the same coefficients with opposite signs. Now we can add and eliminate c:
r+c=78+r−c=52_2r=130 Add the equations
Now, we can easily solve as usual:
2r=130Multiply by the reciprocal of 2r=65Rate of the boat in still water
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
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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.
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.
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.
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?
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?
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?
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?
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?
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?
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?
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.
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.
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
How many grams of pure water must be added to 50 grams of pure acid to make a solution that is 40% acid?
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?
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?
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?
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?
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?
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?
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?
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?