How do you solve #(6x)/(x+4)+4=(2x+2)/(x-1)#?

1 Answer
Dec 23, 2016

Answer:

The solution for this equation is #x =2,-3/2#

Explanation:

#(6x)/(x+4)+4=(2x+2)/(x-1)#

or, #(6x+4*(x+4))/(x+4)=(2x+2)/(x-1) rarr# Take LCM on LHS.

or, #(6x+4x+16)/(x+4)=(2x+2)/(x-1) #

or, #(10x+16)/(x+4)=(2x+2)/(x-1) #

or, #(x-1)(10x+16)=(x+4)(2x+2) rarr# Cross Multiply

or, #x(10x+16)-1(10x+16)=x(2x+2)+4(2x+2) rarr#Simplify

or, #10x^2+16x-10x-16=2x^2+2x+8x+8rarr#Simplify

or, #10x^2+6x-16=2x^2+10x+8#

or, #2(5x^2+3x-8)=2(x^2+5x+4)rarr#Take 2 Common on both LHS Side and RHS Side

or, #(2(5x^2+3x-8))/2=(2(x^2+5x+4))/2rarr#Dividing both sides by 2

or, #5x^2+3x-8=x^2+5x+4#

or, #5x^2-x^2+3x-3x-8=x^2-x^2+5x-3x+4rarr#Subracting #x^2# and 3x from boths sides

or, #4x^2-8=2x+4#

or, #4x^2-8-4=2x+4-4rarr#Subtracting 4 from both sides

or, #4x^2-12=2x#

or, #4x^2-2x-12=2x-2xrarr# Subtracting 2x from both sides

or, #4x^2-2x-12=0#

or, #4x^2-8x+6x-12=0rarr# Factoring LHS

or, #4x(x-2)+6(x-2)=0#

or, #(4x+6)(x-2)=0#

Either 4x+6=0
or, #4x=-6#

or, #x=-(6/4)#

or, #x=-(3/2)#

Or,

#x-2=0#

or, #x=2#

:.The solution for #x=2,-3/2#.