How to find $a+b$ if i know probability function and $\mathsf E[X]$?

Question

How can I find value $a+b$ if the probability function looks the following way: $$ p_X(x)= \left\{ \begin{aligned} &ax+b, \: x\in\{1, 2, 3, 4, 5\}\\ &0, \: \text{otherwise} \end{aligned} \right. $$ and $\mathsf E[X] = 2$

Answer 1

Because $p$ is a probability function: $$\begin{align*}1=\sum p_X(n)&=(a\cdot1+b) + (a\cdot2+b) + (a\cdot3+b) + (a\cdot4+b) + (a\cdot5+b)\\&=15a+5b.\end{align*}$$ Now express that $E[X]=2$: $$\begin{align*}2&=E[X]\\&=\sum n\cdot p_X(n)\\&=1\cdot(a\cdot1+b) + 2\cdot(a\cdot2+b) + 3\cdot(a\cdot3+b) + 4\cdot(a\cdot4+b) + 5\cdot(a\cdot5+b)\\&=55a+15b.\end{align*}$$ So we have to solve the following system of equations: $$\begin{cases}1=15a+5b\\2=55a+15b\end{cases}$$ The solution is: $$\begin{cases}a=-\frac{1}{10}\\b=\frac{1}{2}.\end{cases}$$