Question

graphing an inequality in two variables
graph ( y < \frac{1}{3}x + \frac{1}{2} )
which point is a solution to the inequality?
( circ (-2, 1) )
( circ (-1, 1) )
( circ (0, \frac{1}{2}) )
( circ (3, \frac{1}{2}) )
the correct line is shown.

Explanation:

Step1: Test point (-2,1)

Substitute $x=-2, y=1$ into $y < \frac{1}{3}x + \frac{1}{2}$:
$1 < \frac{1}{3}(-2) + \frac{1}{2} = -\frac{2}{3} + \frac{1}{2} = -\frac{1}{6}$
This is false.

Step2: Test point (-1,1)

Substitute $x=-1, y=1$ into $y < \frac{1}{3}x + \frac{1}{2}$:
$1 < \frac{1}{3}(-1) + \frac{1}{2} = -\frac{1}{3} + \frac{1}{2} = \frac{1}{6}$
This is false.

Step3: Test point $(0, \frac{1}{2})$

Substitute $x=0, y=\frac{1}{2}$ into $y < \frac{1}{3}x + \frac{1}{2}$:
$\frac{1}{2} < \frac{1}{3}(0) + \frac{1}{2} = \frac{1}{2}$
This is false (inequality is strict).

Step4: Test point $(3, \frac{1}{2})$

Substitute $x=3, y=\frac{1}{2}$ into $y < \frac{1}{3}x + \frac{1}{2}$:
$\frac{1}{2} < \frac{1}{3}(3) + \frac{1}{2} = 1 + \frac{1}{2} = \frac{3}{2}$
This is true.

Answer:

D. $(3, \frac{1}{2})$