QUESTION IMAGE
Question
graph each system of linear inequalities.
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Problem 16:
Step 1: Solve the first inequality $-2x < 6$
Divide both sides by $-2$. Remember, when dividing or multiplying an inequality by a negative number, the inequality sign flips.
$\frac{-2x}{-2} > \frac{6}{-2}$
$x > -3$
This is a vertical dashed line (since the inequality is strict, $x > -3$) at $x = -3$, and we shade the region to the right of this line.
Step 2: Graph the second inequality $y \leq 4x + 6$
First, graph the line $y = 4x + 6$. The slope $m = 4$ and the y-intercept $b = 6$. Since the inequality is $\leq$, the line is solid.
To determine the shading, we can test a point not on the line, say $(0, 0)$.
Substitute into the inequality: $0 \leq 4(0) + 6$ which simplifies to $0 \leq 6$, which is true. So we shade the region below (or including) the line $y = 4x + 6$.
Step 3: Find the intersection of the two regions
The solution to the system is the region that is both to the right of $x = -3$ (dashed line) and below (or on) the line $y = 4x + 6$ (solid line).
Problem 17:
Step 1: Graph the first inequality $y \geq \frac{5}{3}x + 2$
Graph the line $y = \frac{5}{3}x + 2$. The slope $m = \frac{5}{3}$ and the y-intercept $b = 2$. Since the inequality is $\geq$, the line is solid.
Test the point $(0, 0)$: $0 \geq \frac{5}{3}(0) + 2$ simplifies to $0 \geq 2$, which is false. So we shade the region above (or including) the line $y = \frac{5}{3}x + 2$.
Step 2: Solve and graph the second inequality $2x - 4y < 4$
First, rewrite it in slope-intercept form ($y = mx + b$):
$-4y < -2x + 4$
Divide both sides by $-4$ (remember to flip the inequality sign):
$y > \frac{1}{2}x - 1$
Graph the line $y = \frac{1}{2}x - 1$ (dashed line, since the inequality is strict). Test the point $(0, 0)$: $0 > \frac{1}{2}(0) - 1$ simplifies to $0 > -1$, which is true. So we shade the region above the dashed line $y = \frac{1}{2}x - 1$.
Step 3: Find the intersection of the two regions
The solution is the region that is above (or on) $y = \frac{5}{3}x + 2$ and above $y = \frac{1}{2}x - 1$.
Problem 18:
Step 1: Solve and graph the first inequality $-8y < 4$
Divide both sides by $-8$ (flip the inequality sign):
$y > -\frac{4}{8}$
$y > -\frac{1}{2}$
This is a horizontal dashed line at $y = -\frac{1}{2}$, and we shade the region above this line.
Step 2: Solve and graph the second inequality $x - 2y \geq 6$
Rewrite in slope-intercept form:
$-2y \geq -x + 6$
Divide by $-2$ (flip the inequality sign):
$y \leq \frac{1}{2}x - 3$
Graph the line $y = \frac{1}{2}x - 3$ (solid line, slope $m = \frac{1}{2}$, y-intercept $b = -3$). Test the point $(0, 0)$: $0 \leq \frac{1}{2}(0) - 3$ simplifies to $0 \leq -3$, which is false. So we shade the region below (or including) the line $y = \frac{1}{2}x - 3$.
Step 3: Find the intersection of the two regions
The solution is the region that is above $y = -\frac{1}{2}$ (dashed line) and below (or on) $y = \frac{1}{2}x - 3$ (solid line).
Problem 19:
Step 1: Solve and graph the first inequality $2x + 4y \geq 12$
Rewrite in slope-intercept form:
$4y \geq -2x + 12$
$y \geq -\frac{1}{2}x + 3$
Graph the line $y = -\frac{1}{2}x + 3$ (solid line, slope $m = -\frac{1}{2}$, y-intercept $b = 3$). Test the point $(0, 0)$: $0 \geq -\frac{1}{2}(0) + 3$ simplifies to $0 \geq 3$, which is false. So we shade the region above (or including) the line $y…
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Problem 16:
Step 1: Solve the first inequality $-2x < 6$
Divide both sides by $-2$. Remember, when dividing or multiplying an inequality by a negative number, the inequality sign flips.
$\frac{-2x}{-2} > \frac{6}{-2}$
$x > -3$
This is a vertical dashed line (since the inequality is strict, $x > -3$) at $x = -3$, and we shade the region to the right of this line.
Step 2: Graph the second inequality $y \leq 4x + 6$
First, graph the line $y = 4x + 6$. The slope $m = 4$ and the y-intercept $b = 6$. Since the inequality is $\leq$, the line is solid.
To determine the shading, we can test a point not on the line, say $(0, 0)$.
Substitute into the inequality: $0 \leq 4(0) + 6$ which simplifies to $0 \leq 6$, which is true. So we shade the region below (or including) the line $y = 4x + 6$.
Step 3: Find the intersection of the two regions
The solution to the system is the region that is both to the right of $x = -3$ (dashed line) and below (or on) the line $y = 4x + 6$ (solid line).
Problem 17:
Step 1: Graph the first inequality $y \geq \frac{5}{3}x + 2$
Graph the line $y = \frac{5}{3}x + 2$. The slope $m = \frac{5}{3}$ and the y-intercept $b = 2$. Since the inequality is $\geq$, the line is solid.
Test the point $(0, 0)$: $0 \geq \frac{5}{3}(0) + 2$ simplifies to $0 \geq 2$, which is false. So we shade the region above (or including) the line $y = \frac{5}{3}x + 2$.
Step 2: Solve and graph the second inequality $2x - 4y < 4$
First, rewrite it in slope-intercept form ($y = mx + b$):
$-4y < -2x + 4$
Divide both sides by $-4$ (remember to flip the inequality sign):
$y > \frac{1}{2}x - 1$
Graph the line $y = \frac{1}{2}x - 1$ (dashed line, since the inequality is strict). Test the point $(0, 0)$: $0 > \frac{1}{2}(0) - 1$ simplifies to $0 > -1$, which is true. So we shade the region above the dashed line $y = \frac{1}{2}x - 1$.
Step 3: Find the intersection of the two regions
The solution is the region that is above (or on) $y = \frac{5}{3}x + 2$ and above $y = \frac{1}{2}x - 1$.
Problem 18:
Step 1: Solve and graph the first inequality $-8y < 4$
Divide both sides by $-8$ (flip the inequality sign):
$y > -\frac{4}{8}$
$y > -\frac{1}{2}$
This is a horizontal dashed line at $y = -\frac{1}{2}$, and we shade the region above this line.
Step 2: Solve and graph the second inequality $x - 2y \geq 6$
Rewrite in slope-intercept form:
$-2y \geq -x + 6$
Divide by $-2$ (flip the inequality sign):
$y \leq \frac{1}{2}x - 3$
Graph the line $y = \frac{1}{2}x - 3$ (solid line, slope $m = \frac{1}{2}$, y-intercept $b = -3$). Test the point $(0, 0)$: $0 \leq \frac{1}{2}(0) - 3$ simplifies to $0 \leq -3$, which is false. So we shade the region below (or including) the line $y = \frac{1}{2}x - 3$.
Step 3: Find the intersection of the two regions
The solution is the region that is above $y = -\frac{1}{2}$ (dashed line) and below (or on) $y = \frac{1}{2}x - 3$ (solid line).
Problem 19:
Step 1: Solve and graph the first inequality $2x + 4y \geq 12$
Rewrite in slope-intercept form:
$4y \geq -2x + 12$
$y \geq -\frac{1}{2}x + 3$
Graph the line $y = -\frac{1}{2}x + 3$ (solid line, slope $m = -\frac{1}{2}$, y-intercept $b = 3$). Test the point $(0, 0)$: $0 \geq -\frac{1}{2}(0) + 3$ simplifies to $0 \geq 3$, which is false. So we shade the region above (or including) the line $y = -\frac{1}{2}x + 3$.
Step 2: Solve and graph the second inequality $6x - 3y < 12$
Rewrite in slope-intercept form:
$-3y < -6x + 12$
Divide by $-3$ (flip the inequality sign):
$y > 2x - 4$
Graph the line $y = 2x - 4$ (dashed line, slope $m = 2$, y-intercept $b = -4$). Test the point $(0, 0)$: $0 > 2(0) - 4$ simplifies to $0 > -4$, which is true. So we shade the region above the dashed line $y = 2x - 4$.
Step 3: Find the intersection of the two regions
The solution is the region that is above (or on) $y = -\frac{1}{2}x + 3$ and above $y = 2x - 4$.
To actually draw these graphs:
- For vertical/horizontal lines, use a straight edge to draw the dashed/solid lines and shade the appropriate regions.
- For the sloped lines, plot the y-intercept, then use the slope (rise over run) to find another point on the line, draw the line, and shade the correct region based on the inequality and the test point.
The final graphs will show the overlapping regions (the solution sets) for each system of inequalities.