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# What two rational expressions sum to 3x−4/x2+x−12?

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### 2 Answers

- stanschimLv 71 month agoFavorite Answer
(3x - 4) / (x^2 + x - 12) = (3x - 4) / [(x + 4)(x - 3)]

(3x - 4) / [(x + 4)(x - 3)] = a / (x + 4) + b / (x - 3)

Multiply through by (x + 4) (x - 3) to give:

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

3x - 4 = ax - 3a + bx + 4b

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

This means that:

3 = a + b, and -4 = -3a + 4b

or

9 = 3a + 3b

-4 = -3a + 4b

Adding gives:

5 = 7b

b = 5/7

a = 3 - (5/7) = 21/7 - 5/7 = 16/7

(3x - 4) / (x^2 + x - 12) = (16/7) / (x + 4) + (5/7) / (x - 3)

- Iggy RockoLv 71 month ago
a/(x + 4) + b/(x - 3) = (3x - 4)/(x^2 + x - 12) for some a and b.

(x - 3)/(x - 3) * a/(x + 4) + (x + 4)/(x + 4) * b/(x - 3) = (3x - 4)/(x^2 + x - 12)

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

ax + bx - 3a + 4b = 3x - 4

solve this system of equations for a and b:

a + b = 3

-3a + 4b = -4

a = 3 - b

-3(3 - b) + 4b = -4

-9 + 3b + 4b = -4

7b = 5

b = 5/7

a = 3 - 5/7 = 16/7

16/(7(x + 4)) + 5/(7(x - 3)) = (3x - 4)/(x^2 + x - 12)