Subtract. Simplify is possible.
(x^2-3x-28)/(x^2-x-42) -(x^2+10x+24)/(x^2+x-12)
factor it ...
(x^2-3x-28)/(x^2-x-42) -(x^2+10x+24)/(x^2+x-12)
= (x-7)(x+4)/((x-7)(x+6)) - (x+6)(x+4)/((x+4)(x-3))
= (x+4)/(x+6) - (x+6)/(x-3)
= [(x+4)(x-3) - (x+6)(x+6)]/((x+6)(x-3))
= ((x^2+x-12)-(x^2+12x+36)]/((x+6)(x-3))
= (-11x - 48)/(x+6)(x-3)
= -(11x+48)/(x+6)(x-3))
check: by using any value of x in original and my answer.
if x=0
original = -28/-42 - 24/-12
= 8/3
my answer = -48/-18 = 8/3
(the probability that my answer is right is quite high)
x^2-3x-28 = (x+3)(x-7)
x^2-x-42 = (x+6)(x-7)
x^2+10x+24 = (x+6)(x+4)
x^2+x-12 = (x-3)(x+4)
so, you are evaluating
(x+3)(x-7)/(x+6)(x-7) - (x+6)(x+4)/(x-3)(x+4)
= (x+3)/(x+6) - (x+6)/(x-3)
LCD is (x+6)(x-3)
[(x+3)(x-3) - (x+6)(x+6)]/(x+6)(x-3)
= (x^2 - 9 - x^2 - 12x - 36)/(x+6)(x-3)
= (-12x - 45)/(x+6)(x-3)
= -3(4x+15)/(x+6)(x-3)
My bad. Reiny is correct. Can you see my mistake?
To subtract and simplify the given expression, we need a common denominator for both fractions. Let's start by finding the factors of the denominators:
For the first fraction's denominator, x^2 - x - 42, we can factor it as (x - 7)(x + 6).
For the second fraction's denominator, x^2 + x - 12, we can factor it as (x - 3)(x + 4).
Now, let's express the fractions with the common denominator:
(x^2 - 3x - 28)/[(x - 7)(x + 6)] - (x^2 + 10x + 24)/[(x - 3)(x + 4)]
To make the denominators the same, we need to multiply the first fraction by (x - 3)/(x - 3) and the second fraction by (x + 6)/(x + 6):
[(x^2 - 3x - 28)(x - 3)]/[(x - 7)(x + 6)(x - 3)] - [(x^2 + 10x + 24)(x + 6)]/[(x - 3)(x + 4)(x + 6)]
Next, we can combine the numerators:
[(x^3 - 6x^2 - 85x - 84) - (x^3 + 16x^2 + 84x + 144)]/[(x - 7)(x + 6)(x - 3)(x + 4)(x + 6)]
Simplifying the numerator:
x^3 - 6x^2 - 85x - 84 - x^3 - 16x^2 - 84x - 144
x^3 - x^3 - 6x^2 - 16x^2 - 85x - 84x - 84 - 144
-22x^2 - 169x - 228
Putting it all together:
[-22x^2 - 169x - 228]/[(x - 7)(x + 6)(x - 3)(x + 4)(x + 6)]
The expression is now subtracted and simplified.