2. Linear Equations and Applications
Chapter 2
Linear Equations and Applications

3. Applications of Linear Equations
2.3
Applications of Linear Equations

4. 2.3 Applications of Linear Equations
Objectives
Translate from words to mathematical expressions.
Write expressions from given information.
Distinguish between expressions and equations.
Use the six steps in solving an applied problem.
Solve percent problems.
Solve investment problems.
Solve mixture problems.
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5. 2.3 Applications of Linear Equations
Problem-Solving Hint
PROBLEM-SOLVING HINT
Usually there are key words and phrases in a verbal problem that translate
into mathematical expressions involving addition, subtraction, multiplication,
and division. Translations of some commonly used expressions follow.
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6. 2.3 Applications of Linear Equations
Translating from Words to Mathematical Expressions
Mathematical Expression
(where x and y are numbers)
Verbal Expression
Addition
The sum of a number and 2
x + 2
3 more than a number
x + 3
7 plus a number
7 + x
16 added to a number
x + 16
A number increased by 9
x + 9
The sum of two numbers
x + y
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7. 2.3 Applications of Linear Equations
Translating from Words to Mathematical Expressions
Mathematical Expression
(where x and y are numbers)
Verbal Expression
Subtraction
4 less than a number
x – 4
10 minus a number
10 – x
A number decreased by 5
x – 5
A number subtracted from 12
12 – x
The difference between two
numbers
x – y
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8. 2.3 Applications of Linear Equations
Translating from Words to Mathematical Expressions
Mathematical Expression
(where x and y are numbers)
Verbal Expression
Multiplication
14 times a number
14x
A number multiplied by 8 8x
of a number (used with
fractions and percent) 3 4 x 3 4
Triple (three times) a number 3x
The product of two numbers xy
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9. 2.3 Applications of Linear Equations
Translating from Words to Mathematical Expressions
Mathematical Expression
(where x and y are numbers)
Verbal Expression
Division
(x ≠ 0) 6 x
The quotient of 6 and a number x 15
A number divided by 15
(y ≠ 0) x y
The ratio of two numbers
or the quotient of two numbers
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10. 2.3 Applications of Linear Equations
Caution
CAUTION
Because subtraction and division are not commutative operations, be careful
to correctly translate expressions involving them. For example, “5 less than a
number” is translated as x – 5, not 5 – x. “A number subtracted from 12” is
expressed as 12 – x, not x – 12.
For division, the number by which we are dividing is the denominator, and
the number into which we are dividing is the numerator. For example, “a
number divided by 15” and “15 divided into x” both translate as Similarly,
“the quotient of x and y” is translated as x 15 x y
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11. 2.3 Applications of Linear Equations
Indicator Words for Equality
Equality
The symbol for equality, =, is often indicated by the word is. In fact, any
words that indicate the idea of “sameness” translate to =.
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12. 2.3 Applications of Linear Equations
Translating Words into Equations
Verbal Sentence
Equation
Twice a number, decreased by 4, is 32.
2x – 4 = 32
If the product of a number and 16 is decreased
by 25, the result is 87.
16x – 25 = 87
x + 6 x
The quotient of a number and the number plus
6 is 48.
= 48 8 x
The quotient of a number and 8, plus the
number, is 54.
+ x = 54
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13. 2.3 Applications of Linear Equations
Distinguishing between Expressions
and Equations
Decide whether each is an expression or an equation.
(a)
4(6 – x) + 2x – 1
There is no equals sign, so this is an expression.
(b)
4(6 – x) + 2x – 1 = –15
Because of the equals sign, this is an equation.
Note that the expression in part (a) simplifies to the expression –2x + 23
and the equation in part (b) has solution 19.
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14. Six Steps to Solving Application Problems
2.3 Applications of Linear Equations
Six Steps to Solving Application Problems
Solving an Applied Problem
Step 1 Read the problem, several times if necessary, until you understand
what is given and what is to be found.
Step 2 Assign a variable to represent the unknown value, using diagrams
or tables as needed. Write down what the variable represents.
Express any other unknown values in terms of the variable.
Step 3 Write an equation using the variable expression(s).
Step 4 Solve the equation.
Step 5 State the answer to the problem. Does it seem reasonable?
Step 6 Check the answer in the words of the original problem.
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15. 2.3 Applications of Linear Equations
Solving a Geometry Problem
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 1
Read the problem. We must find the length and width of the rectangle.
The length is 2 ft more than three times the width and the perimeter is
124 ft.
Step 2
Assign a variable. Let W = the width; then 2 + 3W = length.
Make a sketch. W
2 + 3W
Step 3
Write an equation. The perimeter of a rectangle is given by the
formula P = 2L + 2W.
124 = 2(2 + 3W) + 2W
Let L = 2 + 3W and P = 124.
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16. 2.3 Applications of Linear Equations
Solving a Geometry Problem
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 4
Solve the equation obtained in Step 3.
124 = 2(2 + 3W) + 2W
124 = 4 + 6W + 2W
Distributive property
124 = 4 + 8W
Combine like terms.
124 – 4 = 4 + 8W – 4
Subtract 4.
120 = 8W W =
Divide by 8.
15 = W
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17. 2.3 Applications of Linear Equations
Solving a Geometry Problem
The length of a rectangle is 2 ft more than three times the width. The perimeter
of the rectangle is 124 ft. Find the length and the width of the rectangle.
Step 5
State the answer. The width of the rectangle is 15 ft and the length is
2 + 3(15) = 47 ft.
Step 6
Check the answer by substituting these dimensions into the words of
the original problem.
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18. 2.3 Applications of Linear Equations
Finding Unknown Numerical Quantities
A local grocery store baked a combined total of 912 chocolate chip cookies
and sugar cookies. If they baked 336 more chocolate chip cookies than sugar
cookies, how many of each did the store bake?
Step 1
Read the problem. We are asked to find the total number of each type
of cookie baked.
Step 2
Assign a variable. Let s = the total number of sugar cookies baked.
Since there were 336 more chocolate chip cookies baked,
s = the total number of chocolate chip cookies baked.
Step 3
Write an equation. The sum of the numbers of the cookies baked is
912, so
sugar +
chocolate chip =
total s +
(s + 336) =
912
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19. 2.3 Applications of Linear Equations
Finding Unknown Numerical Quantities
A local grocery store baked a combined total of 912 chocolate chip cookies
and sugar cookies. If they baked 336 more chocolate chip cookies than sugar
cookies, how many of each did the store bake?
Step 4
Solve the equation.
s + (s + 336) = 912
2s = 912
Combine like terms.
2s – 336 = 912 – 336
Subtract 336.
2s = 576 2s =
Divide by 2.
s = 288
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20. 2.3 Applications of Linear Equations
Finding Unknown Numerical Quantities
A local grocery store baked a combined total of 912 chocolate chip cookies
and sugar cookies. If they baked 336 more chocolate chip cookies than sugar
cookies, how many of each did the store bake?
Step 5
State the answer. We let s represent the number of sugar cookies, so
there were 288 sugar cookies baked. Also,
(s + 336) = 624 is the number of chocolate chip cookies baked.
Step 6
Check. 624 is 336 more than 288, and the sum of 624 and 288 is 912.
The conditions of the problem are satisfied, and our answer checks.
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21. 2.3 Applications of Linear Equations
Caution
CAUTION
A common error in solving applied problems is forgetting to answer all the
questions asked in the problem. In the previous example, we were asked to
find the number of each type of cookie baked, so there was an extra step at
the end to find the number of chocolate chip cookies baked.
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22. 2.3 Applications of Linear Equations
Percent Reminder
Percent
Recall from Section 2.2 that percent means “per one hundred,” so 3%
means 0.03, 12% means 0.12, and so on.
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23. 2.3 Applications of Linear Equations
Solving a Percent Problem
During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold
350% more raffle tickets on the second day than the first day, how many raffle
tickets did they sell on the first day?
Step 1
Read the problem. We are given that the number of raffle tickets sold
increased by 350% from the first to the second day, and there were
1440 raffle tickets sold altogether. We must find the number of raffle
tickets sold the first day.
Step 2
Assign a variable. Let x = the total number raffle tickets sold on the
first day of the fundraiser.
350% = 350(.01) = 3.5,
so 3.5x represents the number of raffle tickets sold on the second day
of the fundraiser.
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24. 2.3 Applications of Linear Equations
Solving a Percent Problem
During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold
350% more raffle tickets on the second day than the first day, how many raffle
tickets did they sell on the first day?
Step 3
Write an equation from the information given.
tickets sold the first day +
tickets sold the first day =
total tickets sold x +
(3.5x) =
1440
Step 4
Solve the equation.
1x + 3.5x = 1440
Identity property
4.5x = 1440
Combine like terms.
x = 320
Divide by 4.5.
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25. 2.3 Applications of Linear Equations
Solving a Percent Problem
During a 2-day fundraiser, a local school sold 1440 raffle tickets. If they sold
350% more raffle tickets on the second day than the first day, how many raffle
tickets did they sell on the first day?
Step 5
State the answer. There were 320 raffle tickets sold on the first day
of the 2-day fundraiser.
Step 6
Check that the number of raffle tickets sold on the second day,
1440 – 320 = 1120, is 350% of 320.
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26. 2.3 Applications of Linear Equations
Caution
CAUTION
Avoid two common errors that occur in solving problems like the one in the
previous example.
1. Do not try to find 350% of 1440 and subtract that amount from 1440.
The 350% should be applied to the number of tickets sold on the first day,
not the total number of tickets sold.
2. Do not write the equation as
The percent must be multiplied by some amount; in this case, the amount
is the number of tickets sold on the first day, giving 3.5x.
x = Incorrect
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27. 2.3 Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Step 1
Read the problem. We must find the amount invested in each account.
Step 2
Assign a variable.
The formula for interest is I = p r t.
Let x = the amount to invest at 3%;
50,000 – x = the amount to invest at 5%.
Rate (as a decimal)
Principle
Time
Interest
.03 x 1
0.03x
.05
50,000 – x 1
.05(50,000 – x)
50,000
 Totals 
2180
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28. 2.3 Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Rate (as a decimal)
Interest
0.03
0.03x
Principle
0.05 x
50,000 – x
Time 1
0.05(50,000 – x)
50,000
2180
 Totals 
Step 3
Write an equation. The last column of the table gives the equation.
interest at 3% +
interest at 5% =
total interest
0.03x +
0.05(50,000 – x) =
2180
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29. 2.3 Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Step 4
Solve the equation. We do so without clearing decimals.
0.03x
0.05(50,000 – x) =
2180 +
0.03x (50,000) – 0.05x = 2180
Distributive property
0.03x – 0.05x = 2180
Multiply.
–0.02x = 2180
Combine like terms.
–0.02x = –320
Subtract 2500
x = 16,000
Divide by –.02.
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30. 2.3 Applications of Linear Equations
Solving an Investment Problem
A local company has $50,000 to invest. It will put part of the money in an
account paying 3% interest and the remainder into stocks paying 5%. If the
total annual income from these investments will be $2180, how much will be
invested in each account?
Step 5
State the answer. The company will invest $16,000 at 3%. At 5%, the
company will invest $50,000 – $16,000 = $34,000.
Step 6
Check by finding the annual interest at each rate; they should total
$2180.
0.03($16,000) = $480
and
0.05($34,000) = $1700
$480 + $1700 = $2180, as required.
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31. 2.3 Applications of Linear Equations
Problem-Solving Hint
PROBLEM-SOLVING HINT
In the previous example, we chose to let the variable represent the amount
invested at 3%. Students often ask, “Can I let the variable represent the
other unknown?” The answer is yes. The equation will be different, but in
the end the two answers will be the same.
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32. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 12 L of a 30% acid solution with some 80% solution to get
a 60% solution. How much of the 80% solution should be used?
Step 1
Read the problem. The problem asks for the amount of 80% solution
to be used.
Step 2
Assign a variable. Let x = the number of liters of 80% solution to be
used.
80% + =
30%
80%
30%
12 L
Unknown
number of liters, x
(12 + x)L
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33. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 12 L of a 30% acid solution with some 80% solution to get
a 60% solution. How much of the 80% solution should be used?
Step 2
Assign a variable. Let x = the number of liters of 80% solution
Percent (as a decimal)
Number of Liters
Liters of Pure Acid
0.30 12
0.30(12) = 3.6
0.80 x
0.08x
0.60
12 + x
0.60(12 + x)
Sum must be equal +
80% =
30%
80%
30%
Step 3
Write an equation.
3.6 +
.08x =
.60(12 + x)
12 L
x L
(12 + x)L
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34. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 12 L of a 30% acid solution with some 80% solution to get
a 60% solution. How much of the 80% solution should be used?
Step 4
Solve.
3.6
0.08x
0.60(12 + x) + =
x = x
Distributive property
0.20x = 3.6
Subtract 3.6 and 0.60x.
x = 18
Divide by 0.20.
Step 5
State the answer. The chemist should use 18 L of 80% solution.
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35. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 12 L of a 30% acid solution with some 80% solution to get
a 60% solution. How much of the 80% solution should be used?
Step 6
Check. 12 L of 30% solution plus 18 L of 80% solution is
12(0.30) + 18(0.8) = 18 L
of acid. Similarly, or 30 L of 60% solution has
30(0.6) = 18 L
of acid in the mixture. The total amount of pure acid is 18 L both
before and after mixing, so the answer checks.
Percent (as a decimal)
Number of Liters
Liters of Pure Acid
0.30 12
3.6
0.80 18
14.4
0.60 30 18
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36. 2.3 Applications of Linear Equations
Problem-Solving Hint
PROBLEM-SOLVING HINT
When pure water is added to a solution, remember that water is 0% of the
chemical (acid, alcohol, etc.). Similarly, pure chemical is 100% chemical.
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37. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a
a 40% solution. How much of the pure alcohol solution should be used?
Step 1
Read the problem. The problem asks for the amount of pure alcohol
to be used.
Step 2
Assign a variable. Let x = the number of liters of 100% solution to be
used.
100% + =
10%
100%
10%
8 L
Unknown
number of liters, x
(8 + x)L
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38. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a
a 40% solution. How much of the pure alcohol solution should be used?
Step 2
Assign a variable. Let x = the number of liters of 100% solution
Pure Alcohol (L)
Percent (as a decimal)
Number of Liters
0.10 8
0.10(8) = .8
1.00 x 1x
0.40
8 + x
0.40(8 + x)
Sum must be equal
100% + =
10%
10%
100%
Step 3 .8
Write an equation. + 1x =
.40(8 + x)
8 L
x L
(8 + x)L
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39. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a
a 40% solution. How much of the pure alcohol solution should be used?
Step 4
Solve.
0.8 1x
0.40(8 + x) + =
x = x
Distributive property
0.60x = 2.4
Subtract 0.8 and 0.40x.
x = 4
Divide by 0.60.
Step 5
State the answer. The chemist should use 4 L of pure alcohol.
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40. 2.3 Applications of Linear Equations
Solving a Mixture Problem
A chemist must mix 8 L of a 10% alcohol solution with pure alcohol to get a
a 40% solution. How much of the pure alcohol solution should be used?
Step 6
Check. 8 L of 10% solution plus 4 L pure alcohol is
8(0.10) + 4(1) = 4.8 L
of acid. Similarly, or 12 L of pure alcohol has
12(0.40) = 4.8 L
of alcohol in the mixture. The total amount of pure acid is 4.8 L both
before and after mixing, so the answer checks.
Percent (as a decimal)
Number of Liters
Pure Alcohol (L)
0.10 8
0.8
1.00 4 4
0.40 12
4.8
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