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

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

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

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

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

  1. Point (25, 12):

    • For \( y < 12x \): \[ 12 < 12 \times 25 \quad \text{(True, since 12 < 300)} \]
    • For \( y > 13x + 3 \): \[ 12 > 13 \times 25 + 3 \quad \text{(False, since 12 > 328)} \]
  2. Point (10, 5):

    • For \( y < 12x \): \[ 5 < 12 \times 10 \quad \text{(True, since 5 < 120)} \]
    • For \( y > 13x + 3 \): \[ 5 > 13 \times 10 + 3 \quad \text{(False, since 5 > 133)} \]
  3. Point (2, 2):

    • For \( y < 12x \): \[ 2 < 12 \times 2 \quad \text{(True, since 2 < 24)} \]
    • For \( y > 13x + 3 \): \[ 2 > 13 \times 2 + 3 \quad \text{(False, since 2 > 29)} \]
  4. Point (3, 6):

    • For \( y < 12x \): \[ 6 < 12 \times 3 \quad \text{(True, since 6 < 36)} \]
    • For \( y > 13x + 3 \): \[ 6 > 13 \times 3 + 3 \quad \text{(False, since 6 > 42)} \]

After checking all points, none of the provided points satisfy both inequalities:

  • \( (25, 12) \): satisfies \( y < 12x \) but not \( y > 13x + 3 \)
  • \( (10, 5) \): satisfies \( y < 12x \) but not \( y > 13x + 3 \)
  • \( (2, 2) \): satisfies \( y < 12x \) but not \( y > 13x + 3 \)
  • \( (3, 6) \): satisfies \( y < 12x \) but not \( y > 13x + 3 \)

It appears that there is no valid point among the options given that satisfies both inequalities.