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# I don't know how to answer algebraic word problems.

My teacher didn't explain this to me very well and now I need help because there is a test on Monday.

Hi Charles;
Do you have any homework algebraic word problems you can present us with?
A bill of \$2.35 was paid in dimes and nickels.  If there were 32 coins in all, how many dimes were there?

### 3 Answers by Expert Tutors

Joseph A. | SAT, Math 6-12SAT, Math 6-12
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The main trouble with solving word problems is translating from "English" to "Math." The trouble here could be that you may not realize that there are two equations involved!  In is easy to see that there are 32 coins in total; it was stated as such!  But realizing (and figuring out) the second equation, which represents the coins in terms of their cost, is much harder to see.
4.7 4.7 (7 lesson ratings) (7)
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Let x= number of dimes
Let y = number of nickels

Create a system of 2 equations and solve by substitution or elimination method.

x + y = 32
0.10x + 0.05y = 2.35

Have you learned systems of equations? If so, I will continue showing you how to do this problem.
Amarjeet K. | Professional Engineer for Math and Science TuroringProfessional Engineer for Math and Scien...
4.6 4.6 (8 lesson ratings) (8)
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Let number of dimes be x and number od nickles be y x + y = 32 10x + 5y = 235 Solve simulateous equations x = 32-y substitute in second eqation 10(32-y) + 5y =235 320-10y +5y = 235 320-5y = 235 5y = 320-235 y= 85/5= 17 substitute in first equation x = 15 dimes = 15 and nickles = 17