Ratical and Rational Equations

Question

(4/X^2+X-6)-(1/X^2-4)=(2/X^2+5X+6)

 

Please help me solve

Answer 1

(4/X^2+X-6)-(1/X^2-4)=(2/X^2+5X+6)

Factor all denominators:

4/(x+3)(x-2)  -   1/(x-2)(x+2)  =   2/(x+3)(x+2)

 

Note the common denominator for all three fractions is:  (x+3)(x-2)(x+2)

 

So we multiply all three fractions by this:

 

4(x+3)(x-2)(x+2)/(x+3)(x-2) - 1(x+3)(x-2)(x+2)/(x-2)(x+2) = 2(x+3)(x-2)(x+2)/(x+3)(x+2)

 

And cancel everything we can.....

4(x+3)(x-2)(x+2)/(x+3)(x-2) - 1(x+3)(x-2)(x+2)/(x-2)(x+2) = 2(x+3)(x-2)(x+2)/(x+3)(x+2)

 

Writing down what is left...

4(x+2) - 1(x+3) = 2(x-2)

Then solving....

4x + 8 - x - 3 = 2x -4

3x + 5 = 2x - 4

x = -9

Answer 2

(4/X^2+X-6)-(1/X^2-4)=(2/X^2+5X+6)

Assuming all the x  terms are in the denominator  - factor

 

4/[(X-2)(X+3)]-1/[(X-2)(X+2)]

=2/[(X+2)(X+3)]

 

the common denominator is

[(X-2)(X+3)(X+2)]

 

the domain is restricted

   X!=2, ,-3, -2 ( not equal)

 

4(X+2)-1(X+3)=2(X-2)

 

combine like terms 

X=-9

 

Check!

 

4/[(X-2)(X+3)]-1/[(X-2)(X+2)]
=2/[(X+2)(X+3)]

 

4/[(-11)(-6)]-1/[(-11)(-7)]
=2/[(-7)(-6)]

 

4/66-1/77=2/42

.06-.013=.047

 

Checks out!