To find a solution that satisfies the inequalities \( y < 4x - 2 \) and \( y \leq -\frac{1}{3}x + 2 \), we can look for a point where both conditions are true.
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Graph the inequalities:
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For \( y < 4x - 2 \):
- This is a line with a slope of 4 and a y-intercept of -2. The area below this line represents the solutions for this inequality (since it is a "less than" inequality).
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For \( y \leq -\frac{1}{3}x + 2 \):
- This is a line with a slope of -\(\frac{1}{3}\) and a y-intercept of 2. The area below this line, including the line itself, represents the solutions for this inequality (since it is a "less than or equal to" inequality).
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Find the intersection point: To find where both inequalities are satisfied, we can first find the intersection of the two lines: \[ 4x - 2 = -\frac{1}{3}x + 2 \] Multiply the entire equation by 3 to eliminate the fraction: \[ 12x - 6 = -x + 6 \] Rearranging gives: \[ 12x + x = 6 + 6 \ 13x = 12 \ x = \frac{12}{13} \]
Now, substituting \(x = \frac{12}{13}\) back into either equation to find \(y\): Using \(y = 4x - 2\): \[ y = 4\left(\frac{12}{13}\right) - 2 \ y = \frac{48}{13} - \frac{26}{13} \ y = \frac{22}{13} \]
Thus, the intersection point is \(\left(\frac{12}{13}, \frac{22}{13}\right)\).
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Determine if it's a solution: Now, choose points around this intersection to check which satisfies both inequalities. Let's take the point \(\left(1, 0\right)\) to see if it fits:
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For \(y < 4x - 2\):
- \(0 < 4(1) - 2\)
- \(0 < 2\) (True)
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For \(y \leq -\frac{1}{3}x + 2\):
- \(0 \leq -\frac{1}{3}(1) + 2\)
- \(0 \leq -\frac{1}{3} + 2\)
- \(0 \leq \frac{5}{3}\) (True)
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Thus, the point (1, 0) is a solution to the system of inequalities.
So, one of the solutions would be: \[ \text{(1, 0)} \]