I am given the question:
$y - 2 > -3x$
$x \leq -1 -y$
In the $xy$ plane, if the point with co-ordinates $(a, b)$ lies in the solution set of the system of inequalities above, which of the following relationships between $a$ and $b$ must be true ?
- $a<0$
- $b>0$
- $ab>0$
- $ab<0$
- $a>b$
I was able to determine that $y > -3x + 2$ and $y > -x - 1$
What should be done next ?
You are given
$$y - 2 \gt -3x \tag{1}\label{eq1A}$$
$$x \le -1 -y \tag{2}\label{eq2A}$$
Note your second determined inequality of $y > -x - 1$ is not correct. It should be $y \le -x - 1$ instead.
Multiplying both sides of \eqref{eq2A} by $-3$ gives
$$-3x \ge 3 + 3y \tag{3}\label{eq3A}$$
Combining with \eqref{eq1A} gives
$$y - 2 \gt 3 + 3y \implies 2y \lt -5 \implies y \lt -\frac{5}{2} \tag{4}\label{eq4A}$$
Moving the values to opposite sides (or multiplying both sides by $-1$) in \eqref{eq1A} gives
$$3x \gt -y + 2 \tag{5}\label{eq5A}$$
Next, multiplying sides of \eqref{eq4A} by $-1$ gives
$$-y \gt \frac{5}{2} \tag{6}\label{eq6A}$$
Combining this with \eqref{eq5A} gives
$$3x \gt \frac{5}{2} + 2 \implies x \gt \frac{3}{2} \tag{7}\label{eq7A}$$
Finally, \eqref{eq4A} and \eqref{eq7A} show, with $a = x$ and $b = y$, that statements $1$, $2$ and $3$ are never true, but statements $4$ and $5$ are always true.