SAT Mathematics Practice Test 5

Any point $\left(\mathit{x},\mathit{y}\right)$ that is a solution to the given system of inequalities must satisfy both inequalities in the system. Since the second inequality in the system can be rewritten as the system is equivalent to the following system.$\mathit{y}\le 3\mathit{x}+1$
Since 3x + 1 > x − 1 for $\mathit{x}>-1$ and $3\mathit{x}+1\le \mathit{x}-1$ for$\mathit{x}\le -1$, it follows that for $\mathit{x}>-1$ and $\mathit{y}\le 3\mathit{x}+1$ for $\mathit{x}\le -1$. Of the given choices, only $\left(2,-1\right)$ satisfies these conditions because
Alternate approach: Substituting $\left(2,-1\right)$into the first inequality gives $-1\le 3\left(2\right)+1,$ or $-1\le 7$,which is a true statement. Substituting $\left(2,-1\right)$ into the second inequality gives 2 − (−1) > 1, or 3 > 1, which is a true statement. Therefore, since $\left(2,-1\right)$ satisfies both inequalities, it is a solution to the system.
Choice A is incorrect because substituting −2 for x and −1 for y in the first inequality gives $-1\le 3\left(-2\right)+1,$ or $-1\le -5,$ which is false. Choice B is incorrect because substituting $-1$ for $\mathit{x}$ and $3$ for $\mathit{y}$ in the first inequality gives $3\le 3\left(-1\right)+1,$ or $3\le -2,$ which is false. Choice C is incorrect because substituting 1 for x and 5 for y in the first inequality gives $5\le 3\left(1\right)+1,$ or $5\le 4,$ which is false.