Question

Use the image to answer the question.

A Cartesian coordinate plane's axes range from negative 4 to 4, both by 1-unit increments. 2 dashed lines with arrows at both ends are plotted, and various regions of the graph are marked.

Which point is a solution to the simultaneous inequalities y<12x and y>13x+3 ?

(1 point)
Responses

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

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

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

(3,6)

Answer 1

To determine which point is a solution to the simultaneous inequalities \( y < 12x \) and \( y > 13x + 3 \), we need to plug each point into both inequalities and see which point satisfies both.

1. **Point (25, 12)**:
- For \( y < 12x \):
- \( 12 < 12(25) \) → \( 12 < 300 \) (True)
- For \( y > 13x + 3 \):
- \( 12 > 13(25) + 3 \) → \( 12 > 325 + 3 \) → \( 12 > 328 \) (False)

2. **Point (2, 2)**:
- For \( y < 12x \):
- \( 2 < 12(2) \) → \( 2 < 24 \) (True)
- For \( y > 13x + 3 \):
- \( 2 > 13(2) + 3 \) → \( 2 > 26 + 3 \) → \( 2 > 29 \) (False)

3. **Point (10, 5)**:
- For \( y < 12x \):
- \( 5 < 12(10) \) → \( 5 < 120 \) (True)
- For \( y > 13x + 3 \):
- \( 5 > 13(10) + 3 \) → \( 5 > 130 + 3 \) → \( 5 > 133 \) (False)

4. **Point (3, 6)**:
- For \( y < 12x \):
- \( 6 < 12(3) \) → \( 6 < 36 \) (True)
- For \( y > 13x + 3 \):
- \( 6 > 13(3) + 3 \) → \( 6 > 39 + 3 \) → \( 6 > 42 \) (False)

None of the points given satisfy both inequalities. It appears that there might be an error in the problem, as none of the points provided are valid solutions to both inequalities simultaneously.

If you need a point that can satisfy both, you may need to choose a different point or check the graph for the area where both inequalities hold true.