To determine in which quadrants the solutions for the inequality [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex] exist, we need to analyze the line [tex]\( y = \frac{2}{7} x + 1 \)[/tex], which forms the boundary of the inequality.
First, let's identify the points where the line intersects the axes:
1. Intersection with the x-axis:
- To find the x-intercept, set [tex]\( y = 0 \)[/tex]:
[tex]\[
0 = \frac{2}{7} x + 1
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
\frac{2}{7} x = -1 \implies x = -\frac{7}{2} = -3.5
\][/tex]
2. Intersection with the y-axis:
- To find the y-intercept, set [tex]\( x = 0 \)[/tex]:
[tex]\[
y = \frac{2}{7}(0) + 1 = 1
\][/tex]
Now, plot these intercepts:
- The x-intercept is at [tex]\( (-3.5, 0) \)[/tex].
- The y-intercept is at [tex]\( (0, 1) \)[/tex].
The line [tex]\( y = \frac{2}{7} x + 1 \)[/tex] has a positive slope of [tex]\(\frac{2}{7}\)[/tex], making it relatively shallow. It passes through the second quadrant from the negative x-axis (left side) and through the first quadrant to the positive y-axis (top).
Given this line, we need to determine where the solutions to the inequality [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex] lie. The shaded region representing these solutions will be below or on the line [tex]\( y = \frac{2}{7} x + 1 \)[/tex].
1. First Quadrant (Quadrant I):
- For points in this quadrant, [tex]\( x > 0 \)[/tex] and [tex]\( y > 0 \)[/tex].
- Points in this quadrant can certainly satisfy [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex] as the line continues into this quadrant.
2. Second Quadrant (Quadrant II):
- For points in this quadrant, [tex]\( x < 0 \)[/tex] and [tex]\( y > 0 \)[/tex].
- Points here can also satisfy [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex], as the line exists in this quadrant.
3. Third Quadrant (Quadrant III):
- For points in this quadrant, [tex]\( x < 0 \)[/tex] and [tex]\( y < 0 \)[/tex].
- These points can satisfy [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex] as well, since the line extends beyond the negative y-axis.
4. Fourth Quadrant (Quadrant IV):
- For points in this quadrant, [tex]\( x > 0 \)[/tex] and [tex]\( y < 0 \)[/tex].
- Points here can also satisfy [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex].
Thus, the solutions to the inequality [tex]\( y \leq \frac{2}{7} x + 1 \)[/tex] exist in all four quadrants of the coordinate plane.
Therefore, the answer is:
[tex]\[
\text{All four quadrants}
\][/tex]
