sarah has 16 coins, some quarters and some nickels. if te coins have a value of $2.20, how many of each kind does she have.
quarters ---- x
nickels ----- 20-x
solve for x:
25x + 5(20-x) = 220
typo:
don't know where my 20 came from.
should have been:
quarters ---- x
nickels ----- 16-x
solve for x:
25x + 5(16-x) = 220
To solve this problem, we can set up a system of equations based on the given information.
Let's assume Sarah has x quarters and y nickels.
Based on the given information, we know that:
1. Sarah has a total of 16 coins, so x + y = 16.
2. The total value of the coins is $2.20, so the value of x quarters is 0.25x and the value of y nickels is 0.05y. Therefore, 0.25x + 0.05y = 2.20.
Now, we have a system of two equations:
Equation 1: x + y = 16
Equation 2: 0.25x + 0.05y = 2.20
To solve this system, we can use the substitution method or the elimination method.
Let's solve it using the substitution method:
From Equation 1, we can rewrite it as x = 16 - y.
Substitute x = 16 - y into Equation 2:
0.25(16 - y) + 0.05y = 2.20.
Simplify the equation:
4 - 0.25y + 0.05y = 2.20.
Combine like terms:
-0.20y = -1.80.
Divide both sides by -0.20:
y = -1.80 / -0.20 = 9.
Now, substitute y = 9 back into Equation 1:
x + 9 = 16.
Subtract 9 from both sides:
x = 16 - 9 = 7.
Therefore, Sarah has 7 quarters and 9 nickels.