How do you solve #\frac { x - 6} { x + 2} = \frac { x + 2} { x + 4} + 1#?

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Answer 1 · 2018-05-04T10:52:22

#x=-6#

We can move everything on one side:

#(x-6)/(x+2)-(x+2)/(x+4)-1=0#

From there we have to find the least common multiplier, which will in this case be #(x+2)(x+4)#:

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

From there, we can cross out the mentions on the fractions:

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

Now we can break up the parantheses_

#(x-6)(x+4)=x^2-2x-24#
#-(x+2)(x+2)= -x^2-4x-4#
#-(x+2)(x+4)=-x^2-6x-8#

That will give,
#x^2-2x-24-x^2-4x-4-x^2-6x-8=0#

Which if we shorten, is equal to_
#-x^2-12x-36=0#

If we now use the quadratic equation formula where #a=-1#, #b=-12# and #c=-36#, we will have:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

If we plug in the numbers here we will eventually land on:

#x=-6#