Question

which linear inequality represents the solution set graphed? a 2y + 3x < 10 b 2y + 3x > 10 c 2y - 3x < 10 d 2y - 3x > 10

Explanation:

Step1: Find the equation of the boundary - line

The boundary - line passes through the points $(0,5)$ and $( \frac{10}{3},0)$. The slope $m$ of the line passing through $(x_1,y_1)=(0,5)$ and $(x_2,y_2)=(\frac{10}{3},0)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - 5}{\frac{10}{3}-0}=-\frac{3}{2}$. Using the slope - intercept form $y = mx + b$ with $b = 5$ (the y - intercept), the equation of the line is $y=-\frac{3}{2}x + 5$. Rearranging gives $2y+3x = 10$.

Step2: Determine the inequality sign

The line is dashed, so the inequality is either $2y + 3x<10$ or $2y + 3x>10$. We can test a point in the shaded region, say $(5,5)$. Substitute $x = 5$ and $y = 5$ into the inequalities:
For $2y+3x<10$, we have $2\times5+3\times5=10 + 15=25
ot<10$.
For $2y+3x>10$, we have $2\times5+3\times5=25>10$. So the inequality is $2y + 3x>10$.

Answer:

B. $2y + 3x>10$