I am required to find numbers a and b so that: $$\frac{2x+5}{x^2+x-6}=\frac{a}{x+3}+\frac{b}{x-2}$$
$$\frac{ax-2a\:+\:bx+3b}{x^2+x-6}$$
$$\therefore ax-2a\:+\:bx+3b\:=\:2x+5 $$
To this step i understand my process and the rest i believe would be like a simultaneous equation, or I could even use a trial and error method to find the numbers, but I know there is a much simpler way to solve this, what step is next? or have I made a mistake?
$\endgroup$ 33 Answers
$\begingroup$$$x^2+x-6=(x+3)(x-2)$$
We have $$a(x-2)+b(x+3)=2x+5$$
Set $x-2=0$ and $x+3=0$ one by one
$\endgroup$ 4 $\begingroup$Here's the most basic method. Group the $x$ coefficients $$ (a+b)x + (3b-2a) = 2x + 5 $$
this needs to be true for $\forall x \in \mathbb R$. Therefore
$$ a + b = 2 $$ $$ 3b - 2a = 5 $$
Solve this system for $a,b$
$\endgroup$ 3 $\begingroup$Substitute $x=0$ to get $\frac{a}{3}+\frac{b}{2}=\frac{-5}{6}$ and $x=-1$ to get $\frac{a}{2}+b=\frac{-1}{2}$ and solve them to get $a=-7, b=3$
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