Question

1. $7x+3y\leq18$

$2x-3y\leq9$

Explanation:

Step1: Rewrite to slope-intercept form

For $7x+3y\leq18$:
$3y\leq -7x+18$
$y\leq -\frac{7}{3}x+6$

For $2x-3y\leq9$:
$-3y\leq -2x+9$
$y\geq \frac{2}{3}x-3$

Step2: Identify boundary lines

1. Line 1: $y = -\frac{7}{3}x+6$

• y-intercept: $(0,6)$
• x-intercept: Set $y=0$: $0=-\frac{7}{3}x+6 \implies x=\frac{18}{7}\approx2.57$
• Solid line (due to $\leq$), shade below the line.

1. Line 2: $y = \frac{2}{3}x-3$

• y-intercept: $(0,-3)$
• x-intercept: Set $y=0$: $0=\frac{2}{3}x-3 \implies x=\frac{9}{2}=4.5$
• Solid line (due to $\geq$), shade above the line.

Step3: Find intersection point

Solve

$$\begin{cases} y = -\frac{7}{3}x+6 \\ y = \frac{2}{3}x-3 \end{cases}$$

Set equal: $-\frac{7}{3}x+6 = \frac{2}{3}x-3$
Multiply by 3: $-7x+18 = 2x-9$
$-9x = -27 \implies x=3$
Substitute $x=3$: $y=\frac{2}{3}(3)-3=2-3=-1$
Intersection: $(3,-1)$

Step4: Graph and shade regions

Plot both solid boundary lines, shade below $y=-\frac{7}{3}x+6$ and above $y=\frac{2}{3}x-3$. The overlapping shaded area is the solution set.

Answer:

The solution is the overlapping shaded region bounded by the solid lines $y = -\frac{7}{3}x+6$ (shaded below) and $y = \frac{2}{3}x-3$ (shaded above), with their intersection at $(3, -1)$.