SAT Mathematics Practice Test 5

Any point $(x,y)$ 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 $y < x - 1,$ the system is equivalent to the following system.$y \le 3x + 1$

$y < x - 1$Since 3x + 1 > x − 1 for $x > -1$ and $3x + 1 \le x - 1$ for$x \le -1$, it follows that $y < x - 1$ for $x > -1$ and $y \le 3x + 1$ for $x \le -1$. Of the given choices, only $(2, -1)$ satisfies these conditions because $-1 < 2 - 1 = 1.$ Alternate approach: Substituting $(2,-1)$into the first inequality gives $-1 \le 3(2) + 1,$ or $-1 \le 7$,which is a true statement. Substituting $(2,-1)$ into the second inequality gives 2 − (−1) > 1, or 3 > 1, which is a true statement. Therefore, since $(2,-1)$ 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 $-1 \le 3(-2) + 1,$ or $-1 \le -5,$ which is false. Choice B is incorrect because substituting $-1$ for $x$ and $3$ for $y$ in the first inequality gives $3 \le 3(-1) + 1,$ 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 $5 \le 3(1) + 1,$ or $5 \le 4,$ which is false.