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which point is a solution of the inequality
10x + 12y \leqslant - 20A - (12, -6)B - (-9, 6)C - (9, -4) D - (-12, 8)

1 Answer

3 votes

Given the inequality:

10x + 12y ≤ -20

to check if a point is a solution, you have to replace the point and check if the inequality is satisfied.

A Replacing (12, -6) into the inequality, we get:

10(12) + 12(-6) ≤ -20

120 - 72 ≤ -20

48 ≤ -20

48 is greater than -20, then (12, -6) doesn't satisfy the inequality

B Replacing (-9, 6) into the inequality, we get:

10(-9) + 12(6) ≤ -20

-18 ≤ -20

-18 is greater than -20, then (-9, 6) doesn't satisfy the inequality

C Replacing (9, -4) into the inequality, we get:

10(9) + 12(-4) ≤ -20

42 ≤ -20

42 is greater than -20, then (9, -4) doesn't satisfy the inequality

D Replacing (-12, 8) into the inequality, we get:

10(-12) + 12(8) ≤ -20

-24 ≤ -20

then (-12, 8) satisfies the inequality

User Nick Darvey
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