Help with possible algebra equation My friend has sent me this puzzle, but I’m terrible with Algebra, could someone please explain how to solve it step by step.
For example, how would I find the value of Q?
Puzzle
I wasn’t able to post the picture as my reputation is too low sorry!
 A: It is immediate that $K=7$ and $Q=9$. Then $L+2Z=37$ and $L+Z=24$ yield $Z=13,L=11$. $X$ and $Y$ follow.
A: *

*In the second row, you have $K+K+K+K = 28$, which means $4K = 28 \iff \color{blue}{K = 7}$.

*In the first column, you have $Q+K+K+K = 30$, which means $Q+3K = 30$. Plugging in $K = 7$ yields  $Q+3(7) = 30 \iff \color{green}{Q = 30-21 = 9}$. Alternatively, you may have looked at the third row, which has $2K+2Q = 32$, which gets $Q = 9$ as well.

*In the first row, you have $Q+L+Z+Z = 46$, which means $Q+L+2Z = 46$, and using $Q = 9$, you get $9+L+2Z = 46 \iff \color{purple}{L+2Z = 37} \tag{1} $.

*In the fourth row, you have $K+Z+L+Q = 40$. Using $K = 7$ and $Q = 9$, you get $7+Z+L+9 = 40 \iff \color{purple}{Z+L = 24} \tag{2}$.

*You can solve the system of equations in purple by subtracting the second from the first, eliminating $L$, so you get $\color{red}{Z = 13}$. Plugging it in in either equation yields $\color{brown}{L = 11}$. Hence, all the variables have been found. All that is left is to calculate $X$ and $Y$, which is trivial.

