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y>4x-2 and

y≥-1/3x+2?

3 answers

To analyze the inequalities \( y > 4x - 2 \) and \( y \geq -\frac{1}{3}x + 2 \), we can plot them on a coordinate plane and find the region that satisfies both inequalities.

  1. Equation of the lines:

    • The line for \( y = 4x - 2 \) has a slope of 4 and a y-intercept at \(-2\). Since the inequality is strict (\(>\)), this line will be dashed (not included in the solution).
    • The line for \( y = -\frac{1}{3}x + 2 \) has a slope of \(-\frac{1}{3}\) and a y-intercept at 2. Since the inequality is not strict (\(\geq\)), this line will be solid (included in the solution).

  2. Finding regions:

    • For \( y > 4x - 2 \):
      • The region above the line \( y = 4x - 2 \) is part of the solution.
    • For \( y \geq -\frac{1}{3}x + 2 \):
      • The region above and including the line \( y = -\frac{1}{3}x + 2 \) is part of the solution.

  3. Intersection of the regions:

    • The solution to the system of inequalities will be the area where the region above the dashed line intersects with the area above the solid line.

Steps to sketch the solution:

  1. Draw the line \( y = 4x - 2 \) dashed.
  2. Draw the line \( y = -\frac{1}{3}x + 2 \) solid.
  3. Shade the area above the dashed line for \( y > 4x - 2 \).
  4. Shade the area above and including the solid line for \( y \geq -\frac{1}{3}x + 2 \).
  5. The resulting shaded area where both conditions overlap is the solution set.

Conclusion:

The solution set consists of the region that is above the line \( y = 4x - 2 \) (not including the line) and above or on the line \( y = -\frac{1}{3}x + 2 \). To visualize it, you may want to graph it with appropriate axes.

no so the first one adds up to -1 and the second added up to 5 how?

Let's clarify what you mean by the inequalities and find the values at specific points:

  1. Understanding the inequalities:

    • The first inequality is \( y > 4x - 2 \).
    • The second inequality is \( y \geq -\frac{1}{3}x + 2 \).

  2. Finding intersection points: To find specific points of interest, such as points where these lines intersect or their values at certain values of \( x \), we can set the two equations equal to each other:

    \[ 4x - 2 = -\frac{1}{3}x + 2 \]

    To solve for \( x \):

    \[ 4x + \frac{1}{3}x = 2 + 2 \]

    Multiply through by 3 to eliminate the fraction:

    \[ 12x + x = 6 \] \[ 13x = 6 \implies x = \frac{6}{13} \]

  3. Substituting back to find \( y \): Substitute \( x = \frac{6}{13} \) into either equation to find \( y \):

    Using \( y = 4x - 2 \):

    \[ y = 4\left(\frac{6}{13}\right) - 2 = \frac{24}{13} - 2 = \frac{24}{13} - \frac{26}{13} = -\frac{2}{13} \]

    Now let's check with the second line \( y = -\frac{1}{3}x + 2 \):

    \[ y = -\frac{1}{3}\left(\frac{6}{13}\right) + 2 = -\frac{2}{13} + \frac{26}{13} = \frac{24}{13} \]

In summary, I previously misunderstood what you meant by "first adds up to -1" and "second added up to 5".

To summarize:

  • The intersection point of the two lines is \( \left( \frac{6}{13}, -\frac{2}{13} \right) \).
  • If you want to verify specific values where each of these inequalities are satisfied, or particular combinations that sum to specific values, please provide the details, and I’ll be glad to assist further!