Asked by Javier
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?.
Answers
Answered by
bobpursley
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
.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
Answered by
Quidditch
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.
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.
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