How do you solve 5/(y-3) + 10/(y^2-y-6) = y/(y+2)5y3+10y2y6=yy+2?

1 Answer
Mar 12, 2018

y=10y=10

Explanation:

Move expression to the left side and change its sign

5/(y-3)+10/(y^2-y-6)-y/(y+2)=05y3+10y2y6yy+2=0

Write -yy as a sum or difference

5/(y-3)+10/(y^2+2y-3y-6)-y/(y+2)=05y3+10y2+2y3y6yy+2=0

Factor out yy and -33 from the expression

5/(y-3)+10/(y(y+2)-3(y+2))-y/(y+2)=05y3+10y(y+2)3(y+2)yy+2=0

Factor out y+2y+2 from the expression

5/(y-3)+10/((y+2)(y-3))-y/(y+2)=05y3+10(y+2)(y3)yy+2=0

Write all numerators above the least common denominator

(5(y+2)+10-y(y-3))/((y+2)(y-3))=05(y+2)+10y(y3)(y+2)(y3)=0

Distribute 55 and -yy through the parenthesis

(5y+10+10-y^2+3y)/((y+2)(y-3))=05y+10+10y2+3y(y+2)(y3)=0

Collect the like terms

(8y+20-y^2)/((y+2)(y-3))=08y+20y2(y+2)(y3)=0

Use the commutative property to reorder the terms

(-y^2+8y+20)/((y+2)(y-3))=0y2+8y+20(y+2)(y3)=0

Write 8y8y as a sum or difference

(-y^2+10y-2y+20)/((y+2)(y-3))=0y2+10y2y+20(y+2)(y3)=0

Factor out -yy and -22 from the expression

(-y(y-10)-2(y-10))/((y+2)(y-3))=0y(y10)2(y10)(y+2)(y3)=0

Factor out -(y-10)(y10) from the expression

(-(y-10)(y+2))/((y+2)(y-3))=0(y10)(y+2)(y+2)(y3)=0

Reduce the fraction with y+2y+2

-(y-10)/(y-3)=0y10y3=0

Determine the sign of the fraction

-(y-10)/(y-3)=0y10y3=0

Simplify

(10-y)/(y-3)=010yy3=0

When the quotient of expressions equals 00, the numerator has to be 00

10-y=010y=0

Move the constant, 1010, to the right side and change its sign

-y=-10y=10

Change the signs on both sides of the equation

y=10y=10

Check if the solution is in the defined range

y=10, y!=3,y!=-2y=10,y3,y2

:. y=10