which of the points on the graph is a solution for the stimulus inequalities why is less than 4x - 2 and Y is less than equal to -1/3 x + 2 Enter the x value followed by the Y value

Answer 1

To find a solution that satisfies the inequalities $y<4x-2$ and $y\le -\frac{1}{3}x+2$, we can look for a point where both conditions are true.

1. Graph the inequalities:

   • 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).

   • For $y\le -\frac{1}{3}x+2$:

     • 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).

2. 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: $$12x+x=6+613x=12x=\frac{12}{13}$$

   Now, substituting $x=\frac{12}{13}$ back into either equation to find $y$: Using $y=4x-2$: $$y=4\left(\frac{12}{13}\right)-2y=\frac{48}{13}-\frac{26}{13}y=\frac{22}{13}$$

   Thus, the intersection point is $(\frac{12}{13},\frac{22}{13})$.

3. Determine if it's a solution: Now, choose points around this intersection to check which satisfies both inequalities. Let's take the point $(1,0)$ to see if it fits:

   • For $y<4x-2$:

     • $0<4(1)-2$
     • $0<2$ (True)

   • For $y\le -\frac{1}{3}x+2$:

     • $0\le -\frac{1}{3}(1)+2$
     • $0\le -\frac{1}{3}+2$
     • $0\le \frac{5}{3}$ (True)

Thus, the point (1, 0) is a solution to the system of inequalities.

So, one of the solutions would be: $$\text{(1, 0)}$$