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Which point would not be a solution to the system of linear inequalities shown below?

y, is less than, start fraction, 4, divided by, 5, end fraction, x, minus, 2, y, is greater than, minus, 3, x, plus, 7
y<
5
4
​
x−2y>−3x+7
Answer
Multiple Choice Answers
left parenthesis, 5, comma, minus, 5, right parenthesis(5,−5)
left parenthesis, 10, comma, minus, 8, right parenthesis(10,−8)
left parenthesis, 5, comma, minus, 6, right parenthesis(5,−6)
left parenthesis, minus, 10, comma, minus, 7, right parenthesis(−10,−7)

1 answer

To determine which point does not satisfy the system of inequalities, we need to evaluate each point against both inequalities:

  1. First inequality: \( y < \frac{4}{5}x - 2 \)
  2. Second inequality: \( y > -3x + 7 \)

Now let's evaluate each of the points:

For point (5, -5):

  1. First inequality:
    \(-5 < \frac{4}{5}(5) - 2\)
    \(-5 < 4 - 2\)
    \(-5 < 2\) (True)

  2. Second inequality:
    \(-5 > -3(5) + 7\)
    \(-5 > -15 + 7\)
    \(-5 > -8\) (True)

For point (10, -8):

  1. First inequality:
    \(-8 < \frac{4}{5}(10) - 2\)
    \(-8 < 8 - 2\)
    \(-8 < 6\) (True)

  2. Second inequality:
    \(-8 > -3(10) + 7\)
    \(-8 > -30 + 7\)
    \(-8 > -23\) (True)

For point (5, -6):

  1. First inequality:
    \(-6 < \frac{4}{5}(5) - 2\)
    \(-6 < 4 - 2\)
    \(-6 < 2\) (True)

  2. Second inequality:
    \(-6 > -3(5) + 7\)
    \(-6 > -15 + 7\)
    \(-6 > -8\) (True)

For point (-10, -7):

  1. First inequality:
    \(-7 < \frac{4}{5}(-10) - 2\)
    \(-7 < -8 - 2\)
    \(-7 < -10\) (False)

  2. Second inequality:
    \(-7 > -3(-10) + 7\)
    \(-7 > 30 + 7\)
    \(-7 > 37\) (False)

The point \((-10, -7)\) does not satisfy the first inequality and also fails the second inequality.

Conclusion:

The point that would not be a solution to the system of linear inequalities is \((-10, -7)\).