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# Application of Linear Equations

Author: Michele Harris
##### Description:

The student will be able to use linear equations to solve word problems.

Solving word problems using linear equations.

(more)

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Tutorial

## Problem #1

Justin and his father went fishing and together caught 12 fish.  Three times the number of fish that Justin caught exceeds 12 by as much as 5 times the number that his father caught exceeds 8.  How many fish did each catch?

*start by setting up your notes for Justin and his father.  What do you know about each?

​Justin - "3 times the number of fish that Justin caught exceeds 12"

This means we can represent Justin as:  3x - 12

"by as much as" - this means we set Justin equal to father

​father - "5 times the number his father caught exceeds 8"

This can be represented as:  5(12 - x) - 8

Source: ABEKA Algebra 1 2002

Your equation should be:  3x - 12 = 5 (12 - x) - 8

​Justin = 8

​father = 4

Source: ABEKA Algebra 1 2002

## Problem #2

​In a baseball game 1/2 the number of runs scored by the winning team exceeds 5 by as much as 1/3  the losing team's runs exceeds 1.  If the total runs scored is 18, what is the final score?

*think through the problem carefully, using the same thought process as in the previous problem.

Source: ABEKA Algebra 1 2002

​x = winning team

​18 - x = losing team

​your equation should read:  1/2x - 5 = 1/3 (18 - x) - 1

Source: ABEKA Algebra 1 2002

## Problem #5

A man paid \$8.50 for a small pump and 5 feet of tubing.  He paid 12 times as much for the pump as for each foot of tubing.  How much did the pump cost?  The tubing?

*x = how much he paid for the tubing

​therefore, he paid 5x for the tubing ( bought 5 feet) and 12x for the pump (because he paid 12 times as much as he did for the tubing)

\$8.50 is the total he paid for all

Source: ABEKA Algebra 1 2002

​your equation should look like this:

8.50 = 5x + 12x

​pump = \$6.00

​tubing = \$2.50

Source: ABEKA Algebra 1 2002

## Problem #6

​In lighting a parking lot, a certain number of 500-watt bulbs and twice as many 250-watt bulbs were used.  The total illumination amounted to 5,000 watts.  Find the number of bulbs of each kind used.

​x = number of bulbs

Source: ABEKA Algebra 1 2002

​500x + 2(250x) = 5000

​we multiplied 250x by 2 because there were twice as many bulbs this size.

250-watt = 10

Source: ABEKA Algebra 1 2002

## Problem #7

At the town waterworks, 2 large pumps and 4 smaller ones delivered 4800 gallons of water a minute.  Each of the large pumps delivered 4 times as much water as each small pump.  How many gallons per minute did each pump deliver?

​small = x

​large = 4x

Source: ABEKA Algebra 1 2002

2 (4x) + 4 (x) = 4800

​small = 400 gallons

​large = 1600 gallons

Source: ABEKA Algebra 1 2002

## Problem #8

​At target practice an artillery crew made 11 hits in less than a minute.  If 3/4 of the number of rounds fired was 9 times the number of misses, how many rounds were fired?

​x = fired

​x- 11 = misses

Source: ABEKA Algebra 1 2002

your equation should look like this:

3/4x = 9 (x - 11)

​12 rounds fired

Source: ABEKA Algebra 1 2002

## Problem #9

​A playground is 101 feet longer than it is wide.  If its width were decreased 25 feet, its length would be twice its width.  Find the dimensions of the playground.

w = width

​w + 101 = length

​w - 25 = new width

Source: ABEKA Algebra 1 2002

​w + 101 = 2 (w - 25)

​You are setting the old length equal to the new length

​The dimensions are:  151 ft by 252 ft

Source: ABEKA Algebra 1 2002

## Problem #10

​A man invests 4/5 of his capital at 10 1/2% and the rest at 10%.  His annual income is \$416.  Find his capital.

​x = amount of capital

​4/5x = 10 1/2%

1/5x = 10% ........... 1/5 because that is what is left after taking 4/5 for first percentage

Source: ABEKA Algebra 1 2002

.105 (4/5x) + .10 (1/5x) = 416

​His capital is \$4000

Source: ABEKA Algebra 1 2002

## Problem #14

​In a balloon race, the sum of the distances covered by two of the balloons was 1,025 miles.  If the distance covered by the first balloon was 50 miles more than 1/2 of that covered by the second balloon, how far did each travel?

*d= second balloon

​1/2d + 50 = first balloon

Source: ABEKA Algebra 1 2002

​Your equation should look like this:

​d + 1/2d + 50 = 1025

Source: ABEKA Algebra 1 2002

## Problem #15

A cashier saves 1/5 of her monthly gross salary.  This is \$40 more than is saved by her friend, a secretary, whose monthly gross salary is \$200 greater.  If the secretary saves 1/7 of her salary, find how much each earns per month.

*x = monthly salary for secretary

​x - 200 = cashier

Source: ABEKA Algebra 1 2002