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# Systems of Equations Problems

by Alicia

Solve this problem by using a two order system.

A man bought 42 stamps, some 13¢ and some 18¢. How many of each kind did he buy if the cost was \$6.66?

### Comments for Systems of Equations Problems

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 Mar 19, 2010 Rating Systems of Equations Problems by: Karin Hi Alicia, When you write a system of equations, you have to write two equations. The first thing that you want to ask yourself, is what two different pieces of information to I know? In this problem, you know how many stamps (42) and you know the cost of the stamps. So, we are going to write one equation based on the number of stamps and the other equation based on the cost of the stamps. The next thing you want to do is to define your variables. What pieces of information don't you know that you need to know? You don't know how many 13 cent stamps and how many 18 cent stamps you bought. So, let's define those variables. Let x = the number of 13 cent stamps. Let y = the number of 18 cent stamps. Now let's write our two equations. The number of stamps. x + y = 42 The cost of stamps. .13x + .18y = 6.66 Now that you've written the two equations, we need to solve. You can use substitution or linear combinations. I would use substitution. Let's solve x + y = 42 for y. x -x + y = 42 - x y = -x + 42 Now substitute this into the second equation for y. .13x + .18y = 6.66 .13x + .18(-x+42) = 6.66 - Substitute .13x - .18x + 7.56 = 6.66 - Distribute -.05x + 7.56 = 6.66 - Combine like terms -.05x + 7.56 -7.56 = 6.66 - 7.56 - Subtract 7.56 -.05x = -.90 - Simplify -.05x/-.05 = -.90/-.05 - Divide by -.05 x = 18 X = 18, so there were 18 13 cent stamps. x + y = 42 18 + y = 42 y = 24 There were 24 18 cent stamps. Let's check: .13(18) + .18(24) =6.66 6.66 = 6.66 The final answer is: There were 18 13 cent stamps and 24 18 cent stamps.

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