Please help me solve 3x-2/4-9x^2 - 2x-5/3x^2-13x-10?

Question

I know that the answer is -3x+10/3x^2-13x-10 but I need to know each step to get to that answer. I am especially confused that it is a negative fraction at the end. How do I get that? Please help. 

Answer 1

(3x-2)/(4-9x²) - (2x-5)/(3x²-13x-10)

 

First let's look at the first fraction:

 

numerator:  (3x-2) ...Nothing to factor

 

Denominator:  (4-9x²)

This is the difference of two squares

(4-9x²) = 2²-(3x)² = (2-3x)(2+3x) or (-3x+2)(3x+2)

 

Put this back together:

Fraction 1:  (3x-2)/[(-3x+2)(3x+2)]

 

If we factor -1 from the numerator we get:

 

-(-3x+2)/[(-3x+2)(3x+2)]

Cancelling the (-3x+2) from both we have

 

Fraction 1:  -1/(3x+2)

 

Now look at the 2nd fraction:

 

- (2x-5)/(3x²-13x-10) = (-2x+5)/(3x²-13x-10)

 

Look at the denominator:

We can factor by grouping

Multiply the coefficient of the x² term, 3, by the constant, -10.

3(-10) = -30.  Look for factors of -30 that add to the coefficient

of the x term, -13.  That would be (-15)(2) = -30 and -15 +2 = -13.

Replace -13x with -15x + 2x

 

(3x²-13x-10) = 3x²-15x+2x-10

Group and factor

(3x²-15x)+(2x-10)

(3x)(x-5)+2(x-5)

(3x+2)(x-5)

 

For fraction 2 we now have:

(-2x+5)/[(3x+2)(x-5)]

 

 

-1/(3x+2) + (-2x+5)/[(3x+2)(x-5)]

 

Our common denominator to put this together is:

 

(3x+2)(x-5)

 

-1(x-5)/[(3x+2)(x-5)] + (-2x+5)]/[(3x+2)(x-5)]

 

= (5-x-2x+5)/[(3x+2)(x-5)]

 

= (10-3x)/[(3x+2)(x-5)]

 

Just to verify the denominator:

(3x+2)(x-5) = 3x²-13x-10

which is exactly what you gave as the correct answer.

 

So yes the answer is:

 

(10-3x)/(3x²-13x-10)

Answer 2

Hi! 

Before I start working on this problem, I would just like to know if, for the first half, you mean this: (3x-2)/(4-9x²) or something different. I am pretty sure for the second half of the problem that you have 2x-5 in the numerator and 3x²-13x-10 in the denominator. 

 

I would just like to make sure before taking a crack at it.

Thanks!