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1. which point would not be a solution to the system of linear inequali…

Question

  1. which point would not be a solution to the system of linear inequalities shown below?

$y \leq 3x - 6$ $y \geq -3x - 5$
a. $(6, 9)$ b. $(-9, -1)$
c. $(2, -10)$ d. $(3, 0)$

Explanation:

Step1: Check point A (6,9)

Substitute \(x = 6\), \(y = 9\) into \(y\leq3x - 6\): \(9\leq3(6)-6=12\), true.
Substitute into \(y\geq - 3x - 5\): \(9\geq - 3(6)-5=-23\), true.

Step2: Check point B (-9,-1)

Substitute \(x=-9\), \(y = - 1\) into \(y\leq3x - 6\): \(-1\leq3(-9)-6=-33\), false? Wait, no: \(3(-9)-6=-27 - 6=-33\), \(-1\geq - 33\), wait, no, first inequality: \(y\leq3x - 6\), so \(-1\leq - 33\)? No, that's false? Wait, no, wait second inequality: \(y\geq - 3x - 5\): \(-1\geq - 3(-9)-5=27 - 5=22\)? No, \(-1\geq22\) is false. Wait, maybe I miscalculated. Wait, let's redo B:
First inequality: \(y\leq3x - 6\): \(x=-9\), \(3x - 6=3(-9)-6=-27 - 6=-33\). \(y=-1\), so \(-1\leq - 33\)? No, that's false. Wait, but maybe I messed up. Wait, no, let's check C.

Step3: Check point C (2,-10)

Substitute \(x = 2\), \(y=-10\) into \(y\leq3x - 6\): \(-10\leq3(2)-6=0\), true.
Substitute into \(y\geq - 3x - 5\): \(-10\geq - 3(2)-5=-11\), true (since \(-10 > - 11\)).

Step4: Check point D (3,0)

Substitute \(x = 3\), \(y = 0\) into \(y\leq3x - 6\): \(0\leq3(3)-6=3\), true.
Substitute into \(y\geq - 3x - 5\): \(0\geq - 3(3)-5=-14\), true.

Wait, wait, in Step2, for point B: \(y=-1\), \(3x - 6=3(-9)-6=-33\). So \(-1\leq - 33\) is false. But wait, maybe I made a mistake. Wait, no, let's check again. Wait, the first inequality is \(y\leq3x - 6\), so for B: \(x=-9\), \(3x - 6=-33\), \(y=-1\). So \(-1\leq - 33\) is false. But wait, in Step3, C: \(x=2\), \(3x - 6=0\), \(y=-10\leq0\), true. \(y\geq - 3x - 5\): \(-3(2)-5=-6 - 5=-11\), \(-10\geq - 11\), true.

Wait, but the question is which point is NOT a solution. So we need to find the point that fails at least one inequality.

Wait, let's recheck B:
First inequality: \(y\leq3x - 6\): \(x=-9\), \(3x - 6=-33\), \(y=-1\). So \(-1\leq - 33\)? No, that's false.
Second inequality: \(y\geq - 3x - 5\): \(x=-9\), \(-3x - 5=27 - 5=22\), \(y=-1\geq22\)? No, false.

Wait, but let's check A again: \(x=6\), \(3x - 6=12\), \(y=9\leq12\), true. \(y\geq - 3(6)-5=-23\), \(9\geq - 23\), true.

D: \(x=3\), \(3x - 6=3\), \(y=0\leq3\), true. \(y\geq - 3(3)-5=-14\), \(0\geq - 14\), true.

C: \(x=2\), \(3x - 6=0\), \(y=-10\leq0\), true. \(y\geq - 3(2)-5=-11\), \(-10\geq - 11\), true.

Wait, but B: both inequalities fail? No, wait, maybe I made a mistake with B. Wait, the options: A (6,9), B (-9,-1), C (2,-10), D (3,0). Wait, let's check B again. Wait, first inequality: \(y\leq3x - 6\): \(x=-9\), \(3x - 6=3(-9) - 6=-27 - 6=-33\). \(y=-1\), so \(-1\leq - 33\)? No, that's false. So B fails the first inequality. But wait, maybe the answer is B? Wait, no, wait the question is "which point would NOT be a solution". Wait, but let's check C again. Wait, C: \(x=2\), \(y=-10\). First inequality: \(y\leq3x - 6\): \(-10\leq0\), true. Second inequality: \(y\geq - 3x - 5\): \(-10\geq - 11\), true. So C is a solution. Wait, maybe I messed up B. Wait, no, let's check B's second inequality: \(y\geq - 3x - 5\): \(x=-9\), \(-3x - 5=-3(-9)-5=27 - 5=22\). So \(y=-1\geq22\)? No, false. So B fails both? But that can't be. Wait, maybe the correct answer is B? Wait, no, wait the original problem: let's check again. Wait, maybe I made a mistake with B. Wait, no, let's check the first inequality for B: \(y\leq3x - 6\). \(x=-9\), so \(3*(-9) - 6=-33\). \(y=-1\). So \(-1\leq - 33\) is false. So B is not a solution. But wait, let's check the options again. Wait, maybe the answer is B? Wait, but let's check the other points again. Wait, A: (6,9): \(3*6 - 6=12\), \(9\leq12\), true. \( - 3*6 - 5=-23\), \(9\geq - 23\), true. D: (3,0): \(3*3 - 6=3\),…

Answer:

B. \((-9, -1)\)