SOLUTION: what number must be subtracted from each of the terms of the fractions 13/22 and 17/32 so that the resulting fractions will be the same?

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Question 991533: what number must be subtracted from each of the terms of the fractions 13/22 and 17/32 so that the resulting fractions will be the same?
Found 3 solutions by josgarithmetic, MathTherapy, macston:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
NO SUCH NUMBER!


You wish to find some number r so that 13%2F22-r=17%2F32-r. You have the addition property of equality leaving you no way to solve for r, and no way that the two different given fractions to be made equal from any single number.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

what number must be subtracted from each of the terms of the fractions 13/22 and 17/32 so that the resulting fractions will be the same?
highlight_green%287%29 needs to be subtracted from each numerator and each denominator, if "each of the terms" means the numerator and denominator

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure I understand the question,
and this may get messy.
.
x=number to subtract
.
%2813-x%29%2F%2822-x%29=%2817-x%29%2F%2832-x%29 Cross-multiply
.
%2813-x%29%2832-x%29=%2817-x%29%2822-x%29
.
x%5E2-45x%2B416=x%5E2-39x%2B374
-45x%2B416=-39x%2B374
-6x%2B416=374
-6x=-42
x=7
ANSWER: The number is 7
.
CHECK:
%2813-x%29%2F%2822-x%29=%2817-x%29%2F%2832-x%29
%2813-7%29%2F%2822-7%29=%2817-7%29%2F%2832-7%29
6%2F15=10%2F25
2%2F5=2%2F5