Question

Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Answer 1

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

Add the fractions

$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

Answer 2

Answer:

Step-by-step explanation:

$\frac{6}{y+1}+\frac{y}{y-2}=\frac{6}{y+1} \times \frac{y}{y-2} \\multiply ~by~(y+1)(y-2)\\6(y-2)+y(y+1)=6y\\6y-12+y^2+y=6y\\y^2+y-12=0\\y^2+4y-3y-12=0\\y(y+4)-3(y+4)=0\\(y+4)(y-3)=0\\y=-4,3$