how do i solve this equation? 3y/y+2 + 72/ Y^3+8 = 24/y^2-2y+4algebra

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beckden | High School Teacher | (Level 1) Educator

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3y/(y+2) + 72/(y^3 + 8) = 24/(y^2 - 2y + 4))

1st y^3 + 8 = (y + 2)(y^2 - 2y + 4)   (sum of squares)
so

3y/(y+2) + 72/((y+2)(y^2 - 2y + 4)) = 24/(y^2 - 2y + 4)  
Multiply everythin by (y+2)(y^2-2y+4) and we get

3y(y^2-2y+4) + 72 = 24(y+2) so

3y^3 - 6y^2 + 12y + 72 = 24y + 48, now put in standard form

3y^3 - 6y^2 - 12y + 24 = 0  and we can divide by 3 to get

y^3 - 2y^2 - 4y + 8 = 0   which has -2, 2 as solutions

y^2(y - 2) - 4(y - 2) = 0 so we get

(y-2)(y^2 - 4) = 0 and finally

(y-2)(y-2)(y+2) = 0  so y = -2 or y = 2.

Now this is important, we need to check these solutions.

3y/(y+2) + 72/(y^3+8) = 24/(y^2 - 2y + 4)

When we substitute y = -2 into this equation
3(-2)/(-2+2) + 72/((-2)^3+8) = 24/((-2)^2 - 2(-2) + 4) we get
-6/0 + 72/0 = 24/(12)
we get division by zero so y = -2 is an extraneous solution, but when we substitute y = 2 we get

3(2)/(2+2) + 72/(2^3+8) = 24/(2^2 - 2(2) + 4)

6/4 + 72/16 = 24/(4)
6/4 + 18/4 = 24/4  which is an identity, so
The only solution is y = 2.

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