College algebra help?

Question

4 /( x^2 + 3x - 10) - 1 / (x^2 x -6) = 3 / (x^2 - x -12)

/ = fraction

any and all help is appreciated.

Answer 1

I think you left out a + sign in the second denominator.

Factorise each denominator.  The factorised forms are

(x + 5)(x - 2)

(x + 3)(x - 2)

(x + 3)(x - 4)



If we multiply each of the three terms by (x+5)(x-2)(x+3)(x-4), the denominators will cancel out.

e.g. for the first fraction 4(x+5)(x-2)(x+3)(x-4) / [(x+5)(x-2)]

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

Do this with each of the other fractions, and the equation becomes

4(x+3)(x-4) -1(x+5)(x-4) = 3(x+5)(x-2)

Expand each product:

4(x² - x - 12) - (x² + x - 20) = 3(x² + 3x - 10)

4x² - 4x - 48 - x² - x + 20 = 3x² + 9x - 30

Collect all terms on the left:

-14x + 2 = 0

.......... 2 = 14x

.........2/14 = x

........... x = 1/7

I assumed you knew how to factorise.  Let me know if you need that explained too.

Answer 2

first factor everything out

4/ (x+5)(x-2) - 1/(x+3)(x-2) = 3/ (x-4)(x+3)

lets work on the left side first, so find a common denominator of (x+3)(x+5)(x-2), so you get

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

your new equation looks like this

4(x+3) -(x+5) / (x+5)(x-2)(x+3) = 3/ (x-4)(x+3)

simplify the numerator in the left

4x + 12 - x - 5 = 3x +7

rewrite the equation

(3x+7) / (x+5)(x-2)(x+3) = 3/ (x-4)(x+3)

cross multiply

(3x+7)(x-4)(x+3) = 3(x+5)(x-2)(x+3)

cross out whats on both sides

(3x+7)(x-4) = 3(x+5)(x-2)

factor out both sides of the equation

left: (3x+7)(x-4)= 3x^2 -12x + 7x -28 = 3x^2 -5x -28

right: 3(x+5)(x-2)= 3(x^2 -2x + 5x -10) = 3x^2 +9x -30

new equation looks like

3x^2 -5x -28 = 3x^2 +9x -30

now simplify

14x  = 2

x = 1/7