How do you solve #2/(y+2)+3/y=-y/(y+2)#?

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Answer 1 · 2017-01-05T20:47:46

#x = 1.5 or -6#

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

find LCM of #y# and #y+2#

factors of #y#: #y, 1#

factors of #y+2#: #y+2, 1#

multiply together #y , 1 and (y+2)#:

#y * 1 * (y+2) = y(y+2)#

multiply denominators to get #y(y+2)#:

#(2y)/(y(y+2)) + (3(y+2))/(y(y+2)) = -y^2/(y(y+2))#

multiply everything by #y(y+2)#

hence, #2y + 3(y+2) = -(y^2)#

multiply out brackets:

#2y + 3y + 6 = -(y^2)#

#5y + 6 = -(y^2)#

multiply by #-1#:

#-(5y+6)=y^2#

#6-5y=y^2#

add #5y#:

#y^2+5y = 6#

complete the square:

halve the #y# term #(2.5)#.

square it #(6.25)# and add to both sides.

#5x + 6.25 + x^2 = 6 + 6.25#

#x^2+5x+6.25 = 12.25#

factorise:
#(x + 2.5)(x + 2.5) = 12.25#

calculate the square root of the right side #(3.5 or -3.5)#

#therefore x+2.5 = 3.5 or -3.5#

subtract #2.5#:

#x = 1.5 or -6#