Home > Algebra > Multiplication of Rational Expressions

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

Answer 1

Ethan Adkins

#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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.