Answer:
[tex]\displaystyle a=-\frac{2}{5}\text{ and } b=\frac{2}{5}[/tex]
Step-by-step explanation:
We have the equation:
[tex]\displaystyle \frac{2}{(x+3)(x-2)}=\frac{a}{x+3}+\frac{b}{x-2}[/tex]
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:
[tex]\displaystyle (x+3)(x-2)\frac{2}{(x+3)(x-2)}=(x+3)(x-2)(\frac{a}{x+3}+\frac{b}{x-2})[/tex]
Distribute:
[tex]\displaystyle 2=\frac{(x+3)(x-2)a}{x+3}+\frac{(x+3)(x-2)b}{x-2}[/tex]
Simplify:
[tex]2=a(x-2)+b(x+3)[/tex]
Now, we can determine a and b.
We will let x=2. Then:
[tex]2=a(2-2)+b(2+3)[/tex]
Simplify:
[tex]2=5b[/tex]
So:
[tex]\displaystyle b=\frac{2}{5}[/tex]
Next, we will let x=-3. Then:
[tex]2=a(-3-2)+b(-3+3)[/tex]
Simplify:
[tex]2=-5a[/tex]
So:
[tex]\displaystyle a=-\frac{2}{5}[/tex]
Therefore, our entire equation is:
[tex]\displaystyle \frac{2}{(x+3)(x-2)}=\frac{-\frac{2}{5}}{x+3}+\frac{\frac{2}{5}}{x-2}[/tex]
Or, simpler:
[tex]\displaystyle \frac{2}{(x+3)(x-2)}=-\frac{2}{5(x+3)}+\frac{2}{5(x-2)}[/tex]
