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Algebra Word Problems: Problems with provided values

Word Problems with Multiple Variables and Given Values
This type of problem will be presented such that you'll have to set-up the equation or relation between the variables. Additionally, you will be given the value of one or more variables. On all of these problems you are not asked to solve the problem, only set-up the equation.

Example 1:
“A weather balloon is launched from a height of 100 meters above sea level. The balloon rises at a constant rate of 27 meters per minute. Write an equation that can be used to determine the time in minutes it will take the balloon to reach a height of 2889 meters above sea level.”
Start with the relationship of variables:
Dependent Variable = Fixed Value + Rate of Change * Independent Variable.
“Height of balloon” = “Initial height” + “27meters per minute” h = 100m + 27m
For the final step, substitute the given height of 2889:
2889 = 100 + 27m.

Example 2:
“The perimeter of a rectangle is 40. If the measurement of the length is 3 more than the width. Write an expression that can be used to find the width of the rectangle.”
In this case, you will need to remember the formula for perimeter of a rectangle:
p = 2w + 2l (the same as the sum of its sides: w + w + l + l ).
You will have to substitute the phrase “3 more than the width” in place of length. This expression translates as w + 3.
p = 2w + 2( w + 3).
NOTE: You have to place parenthesis around w + 3 because you’re multiplying 2 by the whole measurement of length. 2w + 3 means “3 more than twice the width”.
Now, you can substitute the known perimeter and you get the final form:
40 = 2w + 2(w + 3)

Example 3:
“Brandon has a budget of $156 to spend on clothes. The shirts he wants to buy are on sale for $24 each, and the pair of pants he wants to buy is $36. Write an expression to help determine the number of shirts Brandon can purchase on his budget.”
Dependent Variable = Rate of Change * Variable + Rate of Change * Variable.
“Total cost”= “24 per shirt” + “36 per pair of pants”
c = 24s + 36p
Because the problem indicates there is only one pair of pants, you substitute and finalize as such:
c = 24s + 36
In budget type problems, when setting up the equations the budget is treated the same way as the total cost because the total cost cannot be more than the budget. Because of this relationship, these problems will be stated as an inequality.
Total cost < budget or Budget > Total Cost
So, substitute the inequality in place of cost and you get:
156 > 24s + 36

Example 4:
“At Turtle South East Book Shop, DVDs on sale cost $8 each, while Blu-Rays cost $14 each. Write an inequality which best describes how many DVDs and Blu-Rays can be purchased for $78 or less.”
Start with relationship of total cost to DVDs and Blu-Rays purchased.
Total Cost = Cost of DVDs + Cost of Blu-Rays
“Total Cost” = “$8 per DVD” + “$14 per Blu-Ray”
c = 8d + 14b.
And finalize by setting up the inequality Budget > Cost.
78 > 8d + 14b