Question

sketch the graph of each linear inequality.

1. $y \geq -3x + 4$
2. $y \leq \frac{3}{5}x - 5$
3. $y > -x - 5$
4. $y > -4$
5. $y > 2x - 5$
6. $y \geq \frac{7}{4}x + 2$

Explanation:

Step1: Graph boundary for 1)

First, plot the line $y=-3x+4$ (solid line, since $\geq$ includes equality). The y-intercept is $(0,4)$, and the x-intercept is when $0=-3x+4 \implies x=\frac{4}{3}\approx1.33$. Shade the region above the line (since $y\geq$ the expression).

Step2: Graph boundary for 2)

Plot the line $y=\frac{3}{5}x-5$ (solid line, since $\leq$ includes equality). The y-intercept is $(0,-5)$, and the x-intercept is when $0=\frac{3}{5}x-5 \implies x=\frac{25}{3}\approx8.33$. Shade the region below the line (since $y\leq$ the expression).

Step3: Graph boundary for 3)

Plot the line $y=-x-5$ (dashed line, since $>$ does not include equality). The y-intercept is $(0,-5)$, and the x-intercept is when $0=-x-5 \implies x=-5$. Shade the region above the line (since $y>$ the expression).

Step4: Graph boundary for 4)

Plot the horizontal line $y=-4$ (dashed line, since $>$ does not include equality). Shade the region above the line (all points where $y$ is greater than -4).

Step5: Graph boundary for 5)

Plot the line $y=2x-5$ (dashed line, since $>$ does not include equality). The y-intercept is $(0,-5)$, and the x-intercept is when $0=2x-5 \implies x=\frac{5}{2}=2.5$. Shade the region above the line (since $y>$ the expression).

Step6: Graph boundary for 6)

Plot the line $y=\frac{7}{4}x+2$ (solid line, since $\geq$ includes equality). The y-intercept is $(0,2)$, and the x-intercept is when $0=\frac{7}{4}x+2 \implies x=-\frac{8}{7}\approx-1.14$. Shade the region above the line (since $y\geq$ the expression).

Answer:

1. Solid line $y=-3x+4$, shade above the line.
2. Solid line $y=\frac{3}{5}x-5$, shade below the line.
3. Dashed line $y=-x-5$, shade above the line.
4. Dashed horizontal line $y=-4$, shade above the line.
5. Dashed line $y=2x-5$, shade above the line.
6. Solid line $y=\frac{7}{4}x+2$, shade above the line.