Answer:
The answer is
[tex]y = 3[/tex]
[tex]y = - 4[/tex]
Step-by-step explanation:
We must find a solution where
[tex] \frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2} [/tex]
Consider the Left Side:
First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get
[tex] \frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1} [/tex]
Which equals
[tex] \frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)} [/tex]
Add the fractions
[tex] \frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2} [/tex]
Simplify the right side by multiplying the fraction
[tex] \frac{6y}{(y + 1)(y + 2)} [/tex]
Set both fractions equal to each other
[tex] \frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)} [/tex]
Since the denomiator are equal, we must set the numerator equal to each other
[tex]6y = {y}^{2} + 7y - 12[/tex]
[tex] = {y}^{2} + y - 12[/tex]
[tex](y + 4)(y - 3)[/tex]
[tex]y = - 4[/tex]
[tex]y = 3[/tex]
Answer:
Step-by-step explanation:
[tex]\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[/tex]
