Like Q12 in LG Ex1-4B - remainder theorem and simultaneous equations

Question: When a polynomial $P(x)$ is divided by $(x+2)(x-4)$ the quotient is the polynomial $Q(x)$ and the remainder is $ax+b$.  Find $a,b$ if $P(-2)=3$ and $P(4)=2$.

Solution.

Using the factor theorem we get two equations for $a,b$ which we can solve to find $a,b$.

We have $$P(x)=(x+2)(x-4)Q(x)+ax+b$$

Therefore

 $P(-2)=3=-2a+b\implies -2a+b=3$  (1)

and

 $P(4)=2=4a+b\implies 4a+b=2$  (2)

Subtract (2)-(2) gives

$-1=6a\implies a=-\dfrac{1}{6}$

Then $b=3+2(-1/6)=2\dfrac{2}{3}=8/3$.

Conclude: $$a=-\dfrac{1}{6}\:\ \mbox{ and }\:\   b=\dfrac{8}{3}$$.

Remember to check my working and tell me if I made an error!! (reward!)