Question

PLEASSSSEEEE ANSWER QUICK

Find values for a and b that makes the following into an identity

`(2)/((x+3)(x-2))=(a)/(x+3)+(b)/(x-2)`

Answer 1

`\displaystyle a=-(2)/(5)\text{ and } b=(2)/(5)`

Explanation:

We have the equation:

`\displaystyle (2)/((x+3)(x-2))=(a)/(x+3)+(b)/(x-2)`

And we want to find a and b such that the equation is true.

First, we can multiply everything by (x+3)(x-2). So:

`\displaystyle (x+3)(x-2)(2)/((x+3)(x-2))=(x+3)(x-2)((a)/(x+3)+(b)/(x-2))`

Distribute:

`\displaystyle 2=((x+3)(x-2)a)/(x+3)+((x+3)(x-2)b)/(x-2)`

Simplify:

`2=a(x-2)+b(x+3)`

Now, we can determine a and b.

We will let x=2. Then:

`2=a(2-2)+b(2+3)`

Simplify:

`2=5b`

So:

`\displaystyle b=(2)/(5)`

Next, we will let x=-3. Then:

`2=a(-3-2)+b(-3+3)`

Simplify:

`2=-5a`

So:

`\displaystyle a=-(2)/(5)`

Therefore, our entire equation is:

`\displaystyle (2)/((x+3)(x-2))=(-(2)/(5))/(x+3)+((2)/(5))/(x-2)`

Or, simpler:

`\displaystyle (2)/((x+3)(x-2))=-(2)/(5(x+3))+(2)/(5(x-2))`