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Sagot :
[tex]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[/tex]
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
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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