1. To graph $4x - 6y \leq -12$, first graph the boundary line $4x - 6y = -12$. Find intercepts: when $x=0$, $-6y=-12$ so $y=2$; when $y=0$, $4x=-12$ so $x=-3$. Plot points (0, 2) and (-3, 0). Since the inequality is $\leq$, draw a solid line. Test point (0,0): $4(0)-6(0)=0 \not\leq -12$, so shade the region not containing (0,0), which is below the line.
2. The graph shows a dashed line with slope $-\frac{2}{3}$ and y-intercept 1, shaded below. The equation is $y = -\frac{2}{3}x + 1$. Since it's dashed and shaded below, the inequality is $y < -\frac{2}{3}x + 1$.
3. The graph shows a solid line with slope $\frac{1}{2}$ and y-intercept -4, shaded above. The line equation is $y = \frac{1}{2}x - 4$. Converting to standard form: $2y = x - 8$ → $x - 2y = 8$ → $5x - 10y = 40$ (multiplying by 5). The shaded region is above the line, so $y \geq \frac{1}{2}x - 4$, which is equivalent to $5x - 10y \leq 20$. Answer: b.
4. To graph $y > \frac{5}{2}x - 3$, graph the boundary line $y = \frac{5}{2}x - 3$. Y-intercept is -3. Slope is $\frac{5}{2}$, so from (0,-3), go up 5, right 2 to (2,2). Draw a dashed line (since >). Shade above the line.
5. To graph $-8x + 9y \geq 72$, graph the boundary line $-8x + 9y = 72$. Find intercepts: when $x=0$, $9y=72$ so $y=8$; when $y=0$, $-8x=72$ so $x=-9$. Plot points (0, 8) and (-9, 0). Since $\geq$, draw a solid line. Test point (0,0): $-8(0)+9(0)=0 \not\geq 72$, so shade the region not containing (0,0), which is above the line.
6. The graph shows a dashed line with slope 1 and y-intercept -1, shaded above. The equation is $y = x - 1$. Since it's dashed and shaded above, the inequality is $y > x - 1$.
7. To graph $y < 3$, draw a horizontal dashed line at $y=3$. Since <, shade below the line.
8. The graph shows a solid vertical line at $x=-1$, shaded to the right. This represents $x \geq -1$. Answer: a.
Parent Tip: Review the logic above to help your child master the concept of linear inequalities worksheet.
