Questions: State the linear inequality whose graph is given in the figure. Write the boundary line equation in the form Ax+By=C, with A, B, and C integers, before stating the inequality. Choose the correct inequality below. A. 4x+3y<-12 B. 4x+3y≤-12 C. 4x+3y>-12

State the linear inequality whose graph is given in the figure. Write the boundary line equation in the form Ax+By=C, with A, B, and C integers, before stating the inequality.

Choose the correct inequality below.
A. 4x+3y<-12
B. 4x+3y≤-12
C. 4x+3y>-12

Transcript text: State the linear inequality whose graph is given in the figure. Write the boundary line equation in the form $A x+B y=C$, with $A, B$, and $C$ integers, before stating the inequality. Choose the correct inequality below. A. $4 x+3 y<-12$ B. $4 x+3 y \leq-12$ C. $4 x+3 y>-12$

Solution

Solution Steps

Step 1: Find the equation of the boundary line.

The boundary line passes through the points $(-6, 4)$ and $(0, -4)$. The slope is $m = \frac{-4 - 4}{0 - (-6)} = \frac{-8}{6} = -\frac{4}{3}$. Using the point-slope form with the point $(0, -4)$, the equation is $y - (-4) = -\frac{4}{3}(x - 0)$, which simplifies to $y + 4 = -\frac{4}{3}x$. Multiplying by 3 gives $3y + 12 = -4x$, or $4x + 3y = -12$.

Step 2: Determine the inequality sign.

The shaded region is below the line, and the line is solid, so the inequality is either $\le$ or $\ge$. We can test the point $(0, 0)$, which is not in the shaded region. Plugging $(0, 0)$ into the equation gives $4(0) + 3(0) = 0$. Since 0 is greater than $-12$, and the shaded region is below the line and does not contain $(0,0)$, the inequality must be $4x + 3y \le -12$.