Question

Which point is a solution to the simultaneous inequalities y < 1/2 x
and y > 1/3 x + 3
?

(1 point)
Responses

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

(25,12)
left parenthesis 25 comma 12 right parenthesis

(10,5)
left parenthesis 10 comma 5 right parenthesis

(3,6)

Answer 1

To find a solution to the simultaneous inequalities \( y < \frac{1}{2}x \) and \( y > \frac{1}{3}x + 3 \), we can substitute each point given into both inequalities.

1. **Point (2, 2)**:
- For \( y < \frac{1}{2}x \):
\( 2 < \frac{1}{2}(2) \rightarrow 2 < 1 \) (FALSE)
- For \( y > \frac{1}{3}x + 3 \):
\( 2 > \frac{1}{3}(2) + 3 \rightarrow 2 > \frac{2}{3} + 3 \rightarrow 2 > \frac{11}{3} \) (FALSE)

2. **Point (25, 12)**:
- For \( y < \frac{1}{2}x \):
\( 12 < \frac{1}{2}(25) \rightarrow 12 < 12.5 \) (TRUE)
- For \( y > \frac{1}{3}x + 3 \):
\( 12 > \frac{1}{3}(25) + 3 \rightarrow 12 > \frac{25}{3} + 3 \rightarrow 12 > \frac{25}{3} + \frac{9}{3} \rightarrow 12 > \frac{34}{3} \) (TRUE)

3. **Point (10, 5)**:
- For \( y < \frac{1}{2}x \):
\( 5 < \frac{1}{2}(10) \rightarrow 5 < 5 \) (FALSE)
- For \( y > \frac{1}{3}x + 3 \):
\( 5 > \frac{1}{3}(10) + 3 \rightarrow 5 > \frac{10}{3} + 3 \rightarrow 5 > \frac{10}{3} + \frac{9}{3} \rightarrow 5 > \frac{19}{3} \) (FALSE)

4. **Point (3, 6)**:
- For \( y < \frac{1}{2}x \):
\( 6 < \frac{1}{2}(3) \rightarrow 6 < 1.5 \) (FALSE)
- For \( y > \frac{1}{3}x + 3 \):
\( 6 > \frac{1}{3}(3) + 3 \rightarrow 6 > 1 + 3 \rightarrow 6 > 4 \) (TRUE)

From our testing, the only point that satisfies both inequalities is:

**(25, 12)**.