Question

algebra i
part b: graph the solution set of the two - variable linear inequality that represents the situation.
$25x + 30ygeq500$

Explanation:

Step1: Find x-intercept (y=0)

Substitute $y=0$ into $25x + 30y \geq 500$:
$25x + 30(0) = 500$
$x = \frac{500}{25} = 20$
Intercept: $(20, 0)$

Step2: Find y-intercept (x=0)

Substitute $x=0$ into $25x + 30y \geq 500$:
$25(0) + 30y = 500$
$y = \frac{500}{30} \approx 16.67$
Intercept: $(0, \frac{500}{30})$

Step3: Rewrite in slope-intercept form

Isolate $y$:
$30y \geq -25x + 500$
$y \geq -\frac{25}{30}x + \frac{500}{30}$
$y \geq -\frac{5}{6}x + \frac{50}{3}$

Step4: Graph the boundary line

Use a solid line (since $\geq$ includes equality) connecting $(20, 0)$ and $(0, \frac{50}{3})$.

Step5: Shade the solution region

Test $(0,0)$: $25(0)+30(0)=0 < 500$, so shade the region above the line (where the inequality holds true). Note: Since the context implies non-negative values for $x$ and $y$ (likely quantities), restrict shading to $x \geq 0, y \geq 0$.

Answer:

1. Draw a solid line connecting the points $(20, 0)$ and $(0, \frac{50}{3}) \approx (0, 16.67)$ on the coordinate plane.
2. Shade the region above this solid line, limited to the first quadrant ($x \geq 0, y \geq 0$), as this is the solution set for $25x + 30y \geq 500$.