Question

Which point is a solution to the simultaneous inequalities y<12x
y
<
1
2
x
and y>13x+3
y
>
1
3
x
+
3
?

(1 point)
Responses

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

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

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

(25,12)

Answer 1

To determine which point is a solution to the simultaneous inequalities \( y < 12x \), \( y < \frac{1}{2}x \), and \( y > 13x + 3 \), we will evaluate each point against these inequalities.

1. **Point (2, 2)**:
\[
y = 2 \quad \text{and} \quad x = 2
\]
- For \( y < 12x \): \( 2 < 12 \cdot 2 \Rightarrow 2 < 24 \) (True)
- For \( y < \frac{1}{2}x \): \( 2 < \frac{1}{2} \cdot 2 \Rightarrow 2 < 1 \) (False)
- For \( y > 13x + 3 \): \( 2 > 13 \cdot 2 + 3 \Rightarrow 2 > 29 \) (False)

Since it fails two inequalities, (2, 2) is not a solution.

2. **Point (3, 6)**:
\[
y = 6 \quad \text{and} \quad x = 3
\]
- For \( y < 12x \): \( 6 < 12 \cdot 3 \Rightarrow 6 < 36 \) (True)
- For \( y < \frac{1}{2}x \): \( 6 < \frac{1}{2} \cdot 3 \Rightarrow 6 < 1.5 \) (False)
- For \( y > 13x + 3 \): \( 6 > 13 \cdot 3 + 3 \Rightarrow 6 > 42 \) (False)

Since it fails two inequalities, (3, 6) is not a solution.

3. **Point (10, 5)**:
\[
y = 5 \quad \text{and} \quad x = 10
\]
- For \( y < 12x \): \( 5 < 12 \cdot 10 \Rightarrow 5 < 120 \) (True)
- For \( y < \frac{1}{2}x \): \( 5 < \frac{1}{2} \cdot 10 \Rightarrow 5 < 5 \) (False)
- For \( y > 13x + 3 \): \( 5 > 13 \cdot 10 + 3 \Rightarrow 5 > 133 \) (False)

Since it fails two inequalities, (10, 5) is not a solution.

4. **Point (25, 12)**:
\[
y = 12 \quad \text{and} \quad x = 25
\]
- For \( y < 12x \): \( 12 < 12 \cdot 25 \Rightarrow 12 < 300 \) (True)
- For \( y < \frac{1}{2}x \): \( 12 < \frac{1}{2} \cdot 25 \Rightarrow 12 < 12.5 \) (True)
- For \( y > 13x + 3 \): \( 12 > 13 \cdot 25 + 3 \Rightarrow 12 > 328 \) (False)

Since it fails one inequality, (25, 12) is not a solution.

Based on evaluating all points against the given inequalities, none of the points provided is a solution to the simultaneous inequalities.

If there was an assumption error in interpreting the inequalities, they can be checked again, but based on the data available, each point is eliminated based on its results in the inequalities.