Mary has 18 coins with a total value of \$2.85. If the coins are dimes and quarters, how many of each type of coins does Mary have

$$x-dimes\\y-quarters\\\\ \left\{\begin{array}{ccc}x+y=18\\0,1x+0.25y=2.85&/\cdot(-10)\end{array}\right.\\+\left\{\begin{array}{ccc}x+y=18\\-x-2.5y=-28.5\end{array}\right\\-\\.\ \ \ \ \ \ -1.5y=-10.5\ \ \ \ /:(-1.5)\\.\ \ \ \ \ \ \ \ \ \ \ \ \ y=7\\\\x+7=18\\x=18-7\\x=11\\\\Answer:11\ dimes\ and\ 7\ quarters$$

Let’s say x represents the number of dimes and y represents the number of quarters.

Note: A dime is worth 10 cents and a quarter is worth 25 cents. Therefore, their coefficients would be 0.1 and 0.25 (you must divide them by 100, since 1 dollar = 100 cents).

Now you can set up a system of equations and solve by substitution.

x + y = 18
0.1x + 0.25y = 2.85

In order to substitute, you have to solve for one of the variables in terms of the other. This will be easiest to do with the first equation.

x + y = 18
y = 18 - x

Now, substitute in that value of y into the second equation.

0.1x + 0.25(18 - x) = 2.85

Now, you can solve for x.

0.1x + 4.5 - 0.25x = 2.85
- 0.15x = -1.65
x = 11

You have 11 dimes.

Now you can substitute that value into your original first equation.

(11) + y = 18
y = 7

You have 7 quarters.

Answer: 11 dimes and 7 quarters

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