When Julia dumped out the big jar of dimes and nickles she found 287 coins. If there was $20.50 in the jar, how many of each kind of coin was there??
Damon answered this question for you here
https://www.jiskha.com/display.cgi?id=1521563768
Why are you reposting it?
I answered a similar question here
Why don't you follow either mine or Damon's method?
https://www.jiskha.com/display.cgi?id=1521562796
I re-posted because @Damon's answer didn't even make sense and the similar question you answered is similar, so I will try that I guess.
To find out how many dimes and nickels Julia had, we can set up an algebraic equation based on the given information.
Let's assume that the number of dimes is represented by "x" and the number of nickels is represented by "y".
We know that the total number of coins is 287, so we can write the equation:
x + y = 287 ---(Equation 1)
We also know that the total value of the coins is $20.50. Since a dime is worth $0.10 and a nickel is worth $0.05, we can write the equation:
0.10x + 0.05y = 20.50 ---(Equation 2)
Now, we have a system of two equations with two unknowns. We can solve this system to find the values of x and y.
First, let's multiply Equation 1 by -0.05 so that we can eliminate the y variable when we add the equations together:
-0.05(x + y) = -0.05(287)
-0.05x - 0.05y = -14.35 ---(Equation 3)
Now, let's add Equations 2 and 3:
(0.10x + 0.05y) + (-0.05x - 0.05y) = 20.50 - 14.35
0.10x - 0.05x + 0.05y - 0.05y = 6.15
0.05x = 6.15
Divide both sides of the equation by 0.05 to isolate x:
x = 6.15 / 0.05
x = 123
Now, substitute the value of x back into Equation 1 to solve for y:
123 + y = 287
y = 287 - 123
y = 164
Therefore, there were 123 dimes and 164 nickels in the jar.