Asked by Ivia
Given the proportion 3x+2y/7=2x+3y/8, determine the value of xy^2-x^2y/x^3+y^3
This is what I got so far...
3x+2y/7=2x+3y/8=k
3x+2y=7k
2x+3y=8k
And that's where I'm stuck at solving this math question.
This is what I got so far...
3x+2y/7=2x+3y/8=k
3x+2y=7k
2x+3y=8k
And that's where I'm stuck at solving this math question.
Answers
Answered by
Steve
(3x+2y)/7 = (2x+3y)/8
8(3x+2y) = 7(2x+3y)
24x+16y = 14x+21y
10x = 5y
2x = y
Using that, we have
(xy^2-x^2y)/(x^3+y^3)
= (x(2x)^2 - x^2(2x))/(x^3+(2x)^3)
= (4x^3-2x^3)/(x^3+8x^3)
= 2/9
8(3x+2y) = 7(2x+3y)
24x+16y = 14x+21y
10x = 5y
2x = y
Using that, we have
(xy^2-x^2y)/(x^3+y^3)
= (x(2x)^2 - x^2(2x))/(x^3+(2x)^3)
= (4x^3-2x^3)/(x^3+8x^3)
= 2/9
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