Question 1129508: Teresa has some dimes and quarters. If she has 28 coins worth a total of
$4.90, how many of each type of coin does she have? Found 2 solutions by josgarithmetic, ikleyn:Answer by josgarithmetic(39621) (Show Source):
Let D be the number of dimes, and let Q be the number of quarters.
Then you have this system of 2 equations in 2 unknowns
D + Q = 28 (1) (there are 28 coins, in all)
10*D + 25*Q = 490 cents (2) (the total)
You can solve the system by different methods.
Let me call some of them: Substitution; Elimination; Cramer's rule (using determinants).
What you choose, depends on your level in Math and on your preferences.
I don't know what your level and preferences are.
So I will choose the most simple - Substitution.
From eq(1) express D = 28-Q and substitute it into eq(2), replacing D there. You will get
10*(28-Q) + 25*Q = 490
280 - 10Q + 25Q = 490
15Q = 490 - 280 = 210
Q = = 14.
Answer. 14 quarters and (the rest) 28-14 = 14 dimes.
Solved.
---------------------
There are different methods for solving such problems:
- using system of two equations (as in this post);
- using single equation in one single unknown;
- and simple logical methods without using equations.
You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.
Read them attentively and become an expert in this field.
(