Sheila brought $1.95 in coins to spend at the bake sale. Her coins were all in dimes and quarters. The number of quarters she had was two less than the number of dimes. How many of each coin did she have?.
q=D-2
.25q+ .10D=1.95
.25(D-2)+.10D=1.95
.25D-.50+.10D=1.95
add .50 to each side.
.35D=2.45
solve for D by dividing each side by .35
then q= D-2
Let d = the number of dimes.
Let q = the number of quarters.
What is known...
d * $0.10 + q * $0.25 = $1.95
and...
q = d - 2
Substitute the above equation for q into the first equation. That gives an equation only in d. Solve for d. Then plug that value into the second equation to find q.
To find out how many quarters and dimes Sheila had, we can set up a system of equations based on the information given.
Let's assume that Sheila had x dimes.
According to the problem, the number of quarters she had was two less than the number of dimes, so the number of quarters would be (x - 2).
Now, let's compute the value of the coins. Since a dime is worth $0.10 and a quarter is worth $0.25, we can form the equation:
0.10x + 0.25(x - 2) = 1.95
Now, we can solve this equation to find the value of x, which represents the number of dimes.