which point is a solution to the simultaneous inequalities y<1/2x and y >1/3x+3
(25,12)
(10,5)
(2,2)
(3,6)
To determine which point is a solution to the simultaneous inequalities, we substitute the coordinates of each point into both inequalities.
For point (25,12):
y < 1/2x
12 < 1/2(25)
12 < 12.5
This inequality is true.
y > 1/3x + 3
12 > 1/3(25) + 3
12 > 8.333 + 3
This inequality is also true.
Since both inequalities are true for point (25,12), it is a solution to the simultaneous inequalities.
For point (10,5):
y < 1/2x
5 < 1/2(10)
5 < 5
This inequality is not true.
y > 1/3x + 3
5 > 1/3(10) + 3
5 > 3.333 + 3
This inequality is true.
Since the first inequality is not true, point (10,5) is not a solution to the simultaneous inequalities.
For point (2,2):
y < 1/2x
2 < 1/2(2)
2 < 1
This inequality is true.
y > 1/3x + 3
2 > 1/3(2) + 3
2 > 0.666 + 3
This inequality is not true.
Since the second inequality is not true, point (2,2) is not a solution to the simultaneous inequalities.
For point (3,6):
y < 1/2x
6 < 1/2(3)
6 < 1.5
This inequality is not true.
y > 1/3x + 3
6 > 1/3(3) + 3
6 > 1 + 3
This inequality is true.
Since the first inequality is not true, point (3,6) is not a solution to the simultaneous inequalities.
Therefore, the only point that is a solution to the simultaneous inequalities is (25,12).