how many 2s and 3s sum to make 15, if there are can only be 6 2s and 3s i was watching a basketball game. a player had scored 15 points. he did this with exactly 6 shots. the shots could only be of 2 or 3 points. i worked out intuitively that he had to shoot 3 3 point shots and 3 2 point shots to get 15 points.
but what would the proper approach be? is there a correct method for this type of thing?
 A: Your first paragraph represents a typical algebraic problem in which you are required to translate word problems from English into mathematical expressions and equations. In the first two sentences, you are given a constraint. The constraint is that the basketball player must score $15$ points with $6$ shots. The shots can take two distinct values - either the basketball player scores $2$ points or the basketball player scores $3$ points.
To rewrite this information into algebraic equations, we can define two variables. Let
$$x=\text{number of 2 point shots}$$
$$y=\text{number of 3 point shots}$$
from our constraint we know that the total number of shots is $6$. Therefore
$$x+y=6$$
and as you are also given that the basketball player scored $15$ points you may conclude this point total is a linear combination of the number of two point shots and the number of three point shots. As a two point shot has a point total of $2$ and a three point shot has a point total of $3$, we see that our second equation is
$$3x+2y = 15$$
We can then rewrite the first equation as $y=6-x$. Substituting this into the second equation forms
$$3x+2(6-x) = 15\implies x+12=15 \implies x=3$$
therefore
$$3+y=6 \implies y=3$$
so we may conclude that the basketball player took three $2$ point shots and three $3$ point shots in order to score $15$ points.
