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

1 Answer
Mar 12, 2018

Answer:

#y=10#

Explanation:

Move expression to the left side and change its sign

#5/(y-3)+10/(y^2-y-6)-y/(y+2)=0#

Write #-y# as a sum or difference

#5/(y-3)+10/(y^2+2y-3y-6)-y/(y+2)=0#

Factor out #y# and #-3# from the expression

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

Factor out #y+2# from the expression

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

Write all numerators above the least common denominator

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

Distribute #5# and #-y# through the parenthesis

#(5y+10+10-y^2+3y)/((y+2)(y-3))=0#

Collect the like terms

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

Use the commutative property to reorder the terms

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

Write #8y# as a sum or difference

#(-y^2+10y-2y+20)/((y+2)(y-3))=0#

Factor out #-y# and #-2# from the expression

#(-y(y-10)-2(y-10))/((y+2)(y-3))=0#

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

#(-(y-10)(y+2))/((y+2)(y-3))=0#

Reduce the fraction with #y+2#

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

Determine the sign of the fraction

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

Simplify

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

When the quotient of expressions equals #0#, the numerator has to be #0#

#10-y=0#

Move the constant, #10#, to the right side and change its sign

#-y=-10#

Change the signs on both sides of the equation

#y=10#

Check if the solution is in the defined range

#y=10, y!=3,y!=-2#

#:. y=10#