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

(3,6)
left parenthesis 3 comma 6 right parenthesis

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

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

(2,2)

Answer 1

To determine which point is a solution to the simultaneous inequalities \(y < 12x\) and \(y > 13x + 3\), we need to substitute each of the proposed points into the inequalities.

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

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

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

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

After evaluating all points, none of them satisfy both inequalities simultaneously. Thus, if you are looking for a point that meets both conditions, there is no solution among the provided options.