1 Applications of Linear Equations. 2 Step 1 Read the problem, several times if necessary, until you understand what is given and what is to be found

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Applicationsof Linear Equations

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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 the 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 your answer. Does it seem reasonable?Step 6 Check your answer in the words of the original

problem.

Six Steps to Solving Applied Problems

Applications of Linear Equations

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A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box?

Example

Read the problem carefully. We are given information about the total number of pens and pencils and asked to find the number of each in the box.

Step 1

Assign a variable.

Let x = the number pencils in the box.

Then x + 16 = the number pens in the box.

Step 2


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Example (continued)

Write an equation.Step 3

The total is the number of pens the number of pencilsplus

68 = ( x + 16 ) + x

Recall: x = # of pencils, x + 16 = # of pens


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Example (continued)

Step 4 Solve the equation.

68 = 2x + 16

68 – 16 = 2x + 16 – 16

52 = 2x

52 2x2 2

=

26 = x

68 = ( x + 16 ) + x

Combine terms.

Subtract 16.

Combine terms.

Divide by 2.

or x = 26


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Example (continued)

Step 5 State the answer. The variable x represents the number of pencils, so there are 26 pencils. Then the number of pens is x + 16 = 26 + 16 = 42.

Recall: x = # of pencils, x + 16 = # of pens

Step 6 Check. Since there are 26 pencils and 42 pens, the combined total number of pencils and pens is 26 + 42 = 68. Because 42 – 26 = 16, there are 16 more pens than pencils. This information agrees with what is given in the problem, so the answer checks.


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A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid,

how many milliliters of water and how many milliliters of acid does it require to fill the beaker?

Example

Read the problem carefully. We must find how many milliliters of water and how many milliliters of acid are needed to fill the beaker.

Step 1

Assign a variable.

Let x = the number of milliliters of acid required.

Then 12x = the number of milliliters of water required.

Step 2


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Example (continued)

Write an equation. A diagram is sometimes helpful.Step 3

Acidx

Water12x

Beaker

= 286

x 12x =+ 286

Recall: x = ml. of acid, 12x = ml. of water.


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Example (continued)

Step 4 Solve.

13x = 286

13x 28613 13

=

x = 22

x + 12x = 286

Combine terms.

Divide by 13.


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Example (continued)

Step 5 State the answer. The beaker requires 22 ml of acid and 12(22) = 264 ml of water.

Recall: x = ml of acid, 12x = ml of water

Step 6 Check. Since 22 + 264 = 286, and 264 is 12 times 22, the answer checks.


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Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized

piece. How long must each piece be?

Example

Read the problem carefully. Three lengths must be found.Step 1

Assign a variable.

x = the length of the middle-sized piece,

3x = the length of the longest piece, and

x – 14 = the length of the shortest piece.

Step 2


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Example (continued)

Step 3 Write an equation.

96 inches

3x x x – 14

Longest Middle-sized Shortest is Total length3x x x – 14 = 96+ +


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Example (continued)

Step 4 Solve.

96 = 5x – 14

96 + 14 = 5x – 14 + 14

110 = 5x

110 5x5 5

=

22 = x

96 = 3x + x + x – 14

Combine terms.

Add 14.

Combine terms.

Divide by 5.

or x = 22


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Example (continued)

Step 5 State the answer. The middle-sized piece is 22 in. long, the longest piece is 3(22) = 66 in. long, and the shortest piece is 22 – 14 = 8 in. long.

Recall: x = length of middle-sized piece, 3x = length of longest piece, and x – 14 = length of shortest piece.

Step 6 Check. The sum of the lengths is 96 in. All conditions of the problem are satisfied.


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“Work problems” are problems that ask you to find long it takes to complete a job when 2 or more people (machines) are working together. In order to solve these

problems use the formula:work = rate of work x time.

Rate of work compares how much can be completed to a period of

time.

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Pipe 1 fills a tank in 6 minutes and pipe 2 fills the same tank in 8

minutes. How long does it take for both pipes together to fill the tank?

• The rate for work for pipe 1 is

• The rate of work for pipe 2 is

• Let x = time together

min 6

tank1

min 8

tank1

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work = rate of work x time

work for pipe 1 = (x)min 6

tank1

work for pipe 2 = (x)min 8

tank1

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To finish, solve the equation

min 7

33or

7

27

247

2434

2418

1

6

124

(job) 18

1

6

1

x

x

xx

xx

xx

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Bob has scored 85, 88, 91, and 86 on his four math tests. He has one

more test left to take. What does he need to score on the last test in order get an A in the class? His

average has to be at least 89.5 to get an A

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Your grade in this class is a "weighted" average which means not all scores count equally. Homework is 7%, quiz average is 10%, tests average is 48% and the final exam is 35% of the final grade. What does a student need to

make on the final exam in order to pass (69.5) if they have the following grades:

HW =80 , QA =72 , and TA =60 ?

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EXAMPLE 7

Solving a Mixture Problem

Step 1 Read the problem. The problem asks for the amount of 80% solution

to be used.

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 to be

used.

+ =30% 80% 30%

80%

12 L Unknown

number of liters, x

(12 + x)L

60%

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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?

• Write an equation.

+ =30% 80%

12 L x (12 + x)L

60%

.30(12) .80 .60(12 )x x

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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?

• Solve the equation.

.30(12) .80 .60(12 )

3.6 .8 7.2 .6

.2 3.6

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x x

x x

x

x

Documents

1 Applications of Linear Equations. 2 Step 1 Read the problem, several times if necessary, until you understand what is given and what is to be found