$x_1 + x_2 + x_3 \le 50$ solutions 
The book shows the answer as attached.
Their equation,
$$x_1 + x_2 + x_3 + y = 50 \implies x_1 + x_2 + x_3 = 50 - y$$
How is that the same as solving,
$$x_1 + x_2 + x_3 \le 50$$
???
 A: It is sometimes called a slack variable. 
If $x_1+x_2+x_3<50$, then there must exist a $y>0$ such that $x_1+x_2+x_3+y = 50$, e.g. if $x_1+x_2+x_3 = 46$ then $y = 4$. If $x_1+x_2+x_3 = 50$ then $y = 0$, so
$$x_1+x_2+x_3 \leq 50 $$
is the same as, for some $y\geq 0$, that
$$x_1+x_2+x_3+y = 50.$$
A: You have $50$ identical candies, and $3$ kids. You want to give $50$ or fewer candies to the $3$ kids (you might choose to distribute all $50$ of the candies, or just $22$, or even (poor kids) a total of $0$ candies. 
We want to know how many ways there are to do the job. You will be possibly keeping some or all of the candies. So in effect you are distributing all  $50$ candies, some to the actual kids, and some to you. Let $x_1$ be the number of candies Kid 1 gets, $x_2$ the number Kid 2 gets, $x_3$ the number Kid 3 gets, and $y$ the number you get. Any distribution of $\le 50$ candies among the $3$ kids leads to a unique solution of $x_1+x_2+x_3+y=50$ in non-negative integers. Conversely, any such solution gives us a unique distribution of $\le 50$ candies among the $3$ kids. So there are exactly as many ways to distribute $\le 50$ candies among $3$ kids as there are to distribute exactly $50$ candies among $4$ "kids."
By standard "Stars and Bars" there are $\binom{53}{3}$ ways to distribute $50$ candies among $4$ "kids."
A: Hint: 
There are now 51 non negative integer possibilities for y, 0 through 50
