For Exercise, determine whether the ordered pair is a solution to the system of inequalities. y < -r + 3 x + 2y < 2 a. (-2, -1) b. (0, -2) c. (0, 1) d. (3, -6)
Given :
y < -x2 + 3 .....(1)
x + 2y ≤ 2 .....(2)
a. (- 2, - 1)
Put x = - 2 and y = - 1 in inequality (1)
- 1 < - (-2)2 + 3
- 1 < - 4 + 3
- 1 < - 1
This is an absurd result. Therefore, (-2, - 1) is not a solution of inequality (1)
Put x = - 2 and y = - 1 in inequality (2)
-2 + 2 (-1) ≤ 2
- 2 - 2 ≤ 2
- 4 ≤ 2
(-2, -1) satisfies inequality (2).
Therefore (- 2, - 1) is not a solution to the system of inequalities as (-2,-1) is not a solution of inequality (1) .
b. (0, - 2)
Put x = 0 and y = - 2 in inequality (1)
- 2 < - (0)2 + 3
- 2 < 0 + 3
- 2 < 3
(0, - 2) is a solution of inequality (1)
Put x = 0 and y = - 2 in inequality (2)
0 + 2 (-2) ≤ 2
- 4 ≤ 2
- 4 ≤ 2
(0, - 2) satisfies inequality (2).
Therefore (0, - 2) is a solution to the system of inequalities.
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