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 – 2*x*. 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 2*x* + 21 – 6*x* = 5. Simplify to get 16 = 4*x*, 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 – 2*x*. 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 2*x* + 3*y* = 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* + 2*x* = 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. |