STEM 698 Algebra Initiative: Word problems that involve systems of equations
1. (CME Algebra I page. 377 adapted) Jeremy helps Madeline build a rectangular pen
with a perimeter of 228 feet. Three times the pen’s width is one less than four times its
length. What are the length and width of the pen?
Let w be the width of the pen in feet and l be the length of the pen in feet.
Then
2w  2l  228

3w  4l  1
This system is equivalent to
2w  2l  228

3w  4l  1
If we multiply the first equation by 2 we get the equivalent system
4w  4l  456

3w  4l  1
Adding the equations, we get 7w  455 , which yields w  65 . Substituting w  65 into
the first equation we get 130  2l  228 , which yields l  49 . Checking in the second
4  49  1  196  1  195 .
equation shows that 3  65  195
So the solution is a width of 65 feet and a length of 49 feet.
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2. Mrs. Smith invested a total of $15000 in two accounts. One account paid 1.5% interest;
the other paid 1.2% interest. In one year she received a total of $190.20 interest. How
many dollars did she invest in each account?
Let x be the amount in dollars invested at 1.5% and y be the amount in dollars invested at
1.2%. Then
 x  y  15000

0.012 x  0.015 y  190.20
The solution is $3400 invested at 1.5% and $11600 invested at 1.2%.
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3. At the mail order company coffeeforless.com, a pound of Lavazza coffee beans costs
$16.00 and a pound of Jamaican Blue Mountain coffee beans costs $112.00. A customer
orders a 5-pound blend of the two brands and pays $252.80. How many pounds of each
brand are in the blend?
3.2 pounds of Lavazza coffee beans and 1.8 pounds of Jamaican Blue Mountain.
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4. A chemist mixes two solutions. One is 24% acid (the rest is water) and the other is
41% acid. About how much of each solution does she need to use to produce 50 gallons
of a solution that is 31% acid? (Express your answer to the nearest hundredth of a
gallon.)
Let x be the amount in gallons of the 24% solution and let y be the amount in gallons of
the 41% acid. Then
 x  y  50

 0.24 x  0.41y
 0.31

x y
This is equivalent to
 x  y  50

0.24 x  0.41y  0.31( x  y)
or
 x  y  50

0.07 x  0.10 y  0
x  500  29.41 and x  350  20.59 .
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There were 29.41 gallons of the 24% solution and 20.59 gallons of the 41% solution.
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