SAT Mathematics Practice Test 7

The first equation can be rewritten as $\mathit{y}\mathit{–}\mathit{x}=3$ and the second as $\frac{\mathit{x}}{4}+\mathit{y}=3$, which implies that $-\mathit{x}=\frac{\mathit{x}}{4}$ , and so $\mathit{x}=0$. The ordered pair (0, 3) satisfies the first equation and also the second, since 0 + 2(3) = 6 is a true equality.
Alternatively, the first equation can be rewritten as $\mathit{y}=\mathit{x}+3$.
Substituting x + 3 for y in the second equation gives $\frac{\mathit{x}}{2}+2\left(\mathit{x}+3\right)=6$.
This can be rewritten using the distributive property as $\frac{\mathit{x}}{2}+2\mathit{x}+6=6$. It follows that $2\mathit{x}{+}_{\mathit{x}}/2$ must be 0. Thus, x = 0. Substituting 0 for x in the equation $\mathit{y}=\mathit{x}+3$ gives y = 3. Therefore, the ordered pair (0, 3) is the solution to the system of equations shown.