We want to solve one equation for either y or x. Since y is all by itself (has a coefficient of 1) in the second equation, that's the easiest variable to solve for. We like our variables like we like our cheese: easy. We solve the second equation for y and get y = 7 – 2x. Next, we can plug this in for y in the first equation: 
Now we have an equation with only an x, so we solve for x. First, distribute the 3 to get 2x + 21 – 6x = 5. Simplify to get 16 = 4x, then simplify further to find x = 4. We have half our point, but we'd still like the other half. Don't rag on us...we won't let it ruin our appetite. We still need to find y. We already solved for y using the second equation, so we know that y = 7 – 2x. Therefore, if x = 4,
We feel fairly certain that the solution to the system of equations is (4, -1). However, it would be better to feel completely certain. We can accomplish that glorious feeling by making sure this solution works in both equations. Is the point (4, -1) a solution to the equation 2x + 3y = 5? When x = 4 and y = -1, the left-hand side of this equation is 2(4) + 3(-1) = 8 – 3, which is indeed 5. Is the point (4, -1) a solution to the equation y + 2x = 7? When x = 4 and y = -1, the left-hand side of this equation is (-1) + 2(4) = -1 + 8, which is indeed 7. These values work in both equations, so the final solution to the system of equations is (4, -1). It better get itself a work visa, because this bad boy is working everywhere. | 
