# decompose fraction

• Feb 27th 2007, 02:07 PM
ceasar_19134
decompose fraction
The fraction:
Code:

```    5x-11     -------     2x^2+x-6```
was obtained byadding the two fractions A/(x+2) and B/(2x-3). Find the value of A + B.
• Feb 27th 2007, 02:40 PM
Jhevon
Quote:

Originally Posted by ceasar_19134
The fraction:
Code:

```    5x-11     -------     2x^2+x-6```
was obtained byadding the two fractions A/(x+2) and B/(2x-3). Find the value of A + B.

So we have (5x - 11)/(2x^2 + x - 6) = A/(x + 2) + B/(2x - 3)

We can go several ways, but they are all amount to the same thing. I'll use the way that requires less typing, tell me if you dont understand something.

Multiply the equation through by (x + 2)(2x - 3), we obtain:

5x - 11 = A(2x - 3) + B(x + 2)
=> 5x - 11 = 2Ax - 3A + Bx + 2B
By grouping like powers of x, we obtain
=> 5x - 11 = (2A + B)x + (2B - 3A)
Now we equate the coefficients of like powers of x on both sides of the equation, we get:

2A + B = 5 .........................(1)
-3A + 2B = -11 ...................(2)

We solve these simultaneous equations to obtain A and B, I think you can take it from here
• Feb 27th 2007, 02:41 PM
Soroban
[size=3]Hello, ceasar_19134![/soze]

Quote:

The fraction:
Code:

```      5x - 11     -----------     2x² + x - 6```
was obtained by adding the two fractions A/(x+2) and B/(2x-3).
Find the value of A + B.

This is a "Partial Fractions" problem.

. . . . . . . . . . . 5x - 11 . . . . . . . .A . . . . . B
We have: . ----------------- . = . -------- + -------
. . . . . . . . (x + 2)(2x - 3) . . . . x + 2 . . 2x - 3

Multiply through by the LCD: (x + 2)(2x - 3)

. . 5x - 11 . = . A(2x - 3) + B(x + 2)

Let x = -2: . -10 - 11 .= .A(-4 - 3) + B(0) . . -21 = -7A . . A = 3

Let x = 3/2: . 15/2 - 11 .= .A(0) + B(3/2 + 2) . . -7/2 = (7/2)B . . B = -1

Therefore: .A + B .= .3 + (-1) .= .2