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The distance a moving body travels, d , varies directly with time, t , it moves. A train travels 100 miles in 2 hours

Write the equation that relates d and t .
How many miles would it travel in 5 hours?

d = 50 t 250 miles

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In the previous example, the variables c and m were named in the problem. Usually that is not the case. We will have to name the variables in the next example as part of the solution, just like we do in most applied problems.

The number of gallons of gas Eunice’s car uses varies directly with the number of miles she drives. Last week she drove 469.8 miles and used 14.5 gallons of gas.

  1. Write the equation that relates the number of gallons of gas used to the number of miles driven.
  2. How many gallons of gas would Eunice’s car use if she drove 1000 miles?

Solution

The number of gallons of gas varies directly with the number of miles driven.
First we will name the variables. Let g = number of gallons of gas.
m = number of miles driven
Write the formula for direct variation. .
We will use g in place of y and m in place of x . .
Substitute the given values for the variables. .
.
Solve for the constant of variation. .
We will round to the nearest thousandth. .
Write the equation that relates g and m . .
Substitute in the constant of variation. .



Find g when m = 1000 . Write the equation that relates g and m . g = 0.031 m Substitute the given value for m . g = 0.031 ( 1000 ) Simplify. g = 31 Eunice’s car would use 31 gallons of gas if she drove it 1,000 miles.

Notice that in this example, the units on the constant of variation are gallons/mile. In everyday life, we usually talk about miles/gallon.

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The distance that Brad travels varies directly with the time spent traveling. Brad travelled 660 miles in 12 hours,

  1. Write the equation that relates the number of miles travelled to the time.
  2. How many miles could Brad travel in 4 hours?

m = 55 h 220 miles

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The weight of a liquid varies directly as its volume. A liquid that weighs 24 pounds has a volume of 4 gallons.

  1. Write the equation that relates the weight to the volume.
  2. If a liquid has volume 13 gallons, what is its weight?

w = 6 v 78 pounds

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In some situations, one variable varies directly with the square of the other variable. When that happens, the equation of direct variation is y = k x 2 . We solve these applications just as we did the previous ones, by substituting the given values into the equation to solve for k .

The maximum load a beam will support varies directly with the square of the diagonal of the beam’s cross-section. A beam with diagonal 4” will support a maximum load of 75 pounds.

  1. Write the equation that relates the maximum load to the cross-section.
  2. What is the maximum load that can be supported by a beam with diagonal 8”?

Solution

The maximum load varies directly with the square of the diagonal of the cross-section.
Name the variables. Let L = maximum load.
c = the diagonal of the cross-section
Write the formula for direct variation, where y varies directly with the square of x . .
We will use L in place of y and c in place of x . .
Substitute the given values for the variables. .
.
Solve for the constant of variation. .
.
Write the equation that relates L and c . .
Substitute in the constant of variation. .



Find L when c = 8 . Write the equation that relates L and c . L = 4.6875 c 2 Substitute the given value for c . L = 4.6875 ( 8 ) 2 Simplify. L = 300 A beam with diagonal 8” could support a maximum load of 300 pounds.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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