Question

consider the system of inequalities,
$-2x + 3y \geq -5$
$9x + 3y < 6$
select all points below which are solutions to the system of inequalities.
$(-1, -1)$
$(-2, -6)$
$(2, -1)$
$(0, 3)$
$(0, 1)$
$(-2, 5)$

Explanation:

Step1: Check point (-1, -1)

Substitute \(x = -1\), \(y = -1\) into \(-2x + 3y \geq -5\):
\(-2(-1) + 3(-1) = 2 - 3 = -1\), and \(-1 \geq -5\) (true).
Substitute into \(9x + 3y < 6\):
\(9(-1) + 3(-1) = -9 - 3 = -12\), and \(-12 < 6\) (true).
So \((-1, -1)\) is a solution.

Step2: Check point (-2, -6)

Substitute \(x = -2\), \(y = -6\) into \(-2x + 3y \geq -5\):
\(-2(-2) + 3(-6) = 4 - 18 = -14\), and \(-14 \geq -5\) (false).
Thus, \((-2, -6)\) is not a solution.

Step3: Check point (2, -1)

Substitute \(x = 2\), \(y = -1\) into \(-2x + 3y \geq -5\):
\(-2(2) + 3(-1) = -4 - 3 = -7\), and \(-7 \geq -5\) (false).
Thus, \((2, -1)\) is not a solution.

Step4: Check point (0, 3)

Substitute \(x = 0\), \(y = 3\) into \(9x + 3y < 6\):
\(9(0) + 3(3) = 9\), and \(9 < 6\) (false).
Thus, \((0, 3)\) is not a solution.

Step5: Check point (0, 1)

Substitute \(x = 0\), \(y = 1\) into \(-2x + 3y \geq -5\):
\(-2(0) + 3(1) = 3\), and \(3 \geq -5\) (true).
Substitute into \(9x + 3y < 6\):
\(9(0) + 3(1) = 3\), and \(3 < 6\) (true).
So \((0, 1)\) is a solution.

Step6: Check point (-2, 5)

Substitute \(x = -2\), \(y = 5\) into \(9x + 3y < 6\):
\(9(-2) + 3(5) = -18 + 15 = -3\), and \(-3 < 6\) (true).
Substitute into \(-2x + 3y \geq -5\):
\(-2(-2) + 3(5) = 4 + 15 = 19\), and \(19 \geq -5\) (true).
So \((-2, 5)\) is a solution.

Answer:

\((-1, -1)\), \((0, 1)\), \((-2, 5)\)