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2-Variable Equations & Inequalities Online Practice

Complete this assessment to review what you’ve learned. It will not count toward your grade.
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Question
Use the image to answer the question.

Two dashed lines and four points are graphed on a coordinate plane. The x-axis ranges from negative 5 to 15 in increments of 1. The y-axis ranges from negative 10 to 12 in increments of 1.

Which point is a solution for the simultaneous inequalities y>−2x+3
and y<3x−3
?

(1 point)
Responses

(2,5)
left parenthesis 2 comma 5 right parenthesis

(3,6)
left parenthesis 3 comma 6 right parenthesis

(1,−5)
left parenthesis 1 comma negative 5 right parenthesis

(10,5)
left parenthesis 10 comma 5 right parenthesis
Remaining Attempts : 3
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1 answer

To find a point that is a solution for the simultaneous inequalities \( y > -2x + 3 \) and \( y < 3x - 3 \), we can evaluate each of the provided points by substituting their coordinates into the inequalities.

  1. For the point (2, 5):

    • Check \( y > -2x + 3 \): \( 5 > -2(2) + 3 \) \( 5 > -4 + 3 \) \( 5 > -1 \) (True)
    • Check \( y < 3x - 3 \): \( 5 < 3(2) - 3 \) \( 5 < 6 - 3 \) \( 5 < 3 \) (False)

  2. For the point (3, 6):

    • Check \( y > -2x + 3 \): \( 6 > -2(3) + 3 \) \( 6 > -6 + 3 \) \( 6 > -3 \) (True)
    • Check \( y < 3x - 3 \): \( 6 < 3(3) - 3 \) \( 6 < 9 - 3 \) \( 6 < 6 \) (False)

  3. For the point (1, -5):

    • Check \( y > -2x + 3 \): \( -5 > -2(1) + 3 \) \( -5 > -2 + 3 \) \( -5 > 1 \) (False)
    • Check \( y < 3x - 3 \): \( -5 < 3(1) - 3 \) \( -5 < 3 - 3 \) \( -5 < 0 \) (True)

  4. For the point (10, 5):

    • Check \( y > -2x + 3 \): \( 5 > -2(10) + 3 \) \( 5 > -20 + 3 \) \( 5 > -17 \) (True)
    • Check \( y < 3x - 3 \): \( 5 < 3(10) - 3 \) \( 5 < 30 - 3 \) \( 5 < 27 \) (True)

Now, summarizing the results:

  • (2, 5): False for the second inequality
  • (3, 6): False for the second inequality
  • (1, -5): False for the first inequality
  • (10, 5): True for both inequalities

Thus, the point that is a solution for both inequalities is (10, 5).