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To solve the system of linear inequalities [tex]\( 2x - 3y \leq 12 \)[/tex] and [tex]\( y < -3 \)[/tex], we need to determine the region of the coordinate plane that satisfies both inequalities. Let's examine each inequality separately and then combine the results.
### Step-by-Step Solution:
1. Rewrite each inequality in a more convenient form:
- For [tex]\( 2x - 3y \leq 12 \)[/tex]:
- Let's rewrite it in slope-intercept form [tex]\( y = mx + b \)[/tex].
- First, solve for [tex]\( y \)[/tex]:
[tex]\[ 2x - 3y \leq 12 \implies -3y \leq 12 - 2x \implies y \geq \frac{2x - 12}{3} \][/tex]
- This means the inequality represents a region on or above the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex].
- For [tex]\( y < -3 \)[/tex]:
- This simply means the region below the horizontal line [tex]\( y = -3 \)[/tex].
2. Graph each inequality:
- Graph [tex]\( y \geq \frac{2x}{3} - 4 \)[/tex]:
- Plot the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex].
- Since it is [tex]\( \geq \)[/tex], we shade the region above this line.
[tex]\[ \begin{array}{c|c} x & y = \frac{2x}{3} - 4 \\ \hline 0 & -4 \\ 3 & -2 \\ -3 & -6 \\ \end{array} \][/tex]
Plotting these points on the graph and drawing the line connecting them.
- Graph [tex]\( y < -3 \)[/tex]:
- Draw a dashed horizontal line at [tex]\( y = -3 \)[/tex] to indicate that [tex]\( y \)[/tex] is strictly less than [tex]\(-3\)[/tex].
- Shade the region below this line.
3. Combine the shaded regions:
- The solution to the system of inequalities is the intersection of the regions shaded for each inequality.
- Thus, we are looking for the region where:
- [tex]\( y \geq \frac{2x}{3} - 4 \)[/tex]
- And [tex]\( y < -3 \)[/tex]
4. Determine the intersection visually or via algebraic approaches:
- We need the [tex]\( y \)[/tex]-values that simultaneously satisfy both conditions.
- Clearly, [tex]\( y \)[/tex] must be less than [tex]\(-3\)[/tex], and it must be on or above [tex]\( \frac{2x}{3} - 4 \)[/tex].
5. Note the boundaries and solution region:
- From plotting, the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex] intersects [tex]\( y < -3 \)[/tex]. We substitute [tex]\( y = -3 \)[/tex] into [tex]\( 2x - 3y = 12 \)[/tex]:
[tex]\[ 2x - 3(-3) = 12 \implies 2x + 9 = 12 \implies 2x = 3 \implies x = \frac{3}{2} \][/tex]
- Thus, at [tex]\( x = \frac{3}{2} \)[/tex] on the line [tex]\( y = -3 \)[/tex], combine this with checking [tex]\( y \)[/tex]-value condition:
- Any point [tex]\( (x, -3) \)[/tex], for [tex]\( x \leq \frac{3}{2} \)[/tex].
### Final Graph Conclusion:
- The final graph shows that the solution region is bounded by:
- The [tex]\( y = \frac{2x}{3} - 4 \)[/tex] line (region above it).
- [tex]\( y = -3 \)[/tex] line (strictly below it).
- And the intersection from [tex]\( x = -\infty \)[/tex] to [tex]\( x = \frac{3}{2} \)[/tex].
Thus, the graph of your solution appears as the region above the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex] but strictly below [tex]\( y = -3 \)[/tex] extending horizontally where conditions intersect.
### Step-by-Step Solution:
1. Rewrite each inequality in a more convenient form:
- For [tex]\( 2x - 3y \leq 12 \)[/tex]:
- Let's rewrite it in slope-intercept form [tex]\( y = mx + b \)[/tex].
- First, solve for [tex]\( y \)[/tex]:
[tex]\[ 2x - 3y \leq 12 \implies -3y \leq 12 - 2x \implies y \geq \frac{2x - 12}{3} \][/tex]
- This means the inequality represents a region on or above the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex].
- For [tex]\( y < -3 \)[/tex]:
- This simply means the region below the horizontal line [tex]\( y = -3 \)[/tex].
2. Graph each inequality:
- Graph [tex]\( y \geq \frac{2x}{3} - 4 \)[/tex]:
- Plot the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex].
- Since it is [tex]\( \geq \)[/tex], we shade the region above this line.
[tex]\[ \begin{array}{c|c} x & y = \frac{2x}{3} - 4 \\ \hline 0 & -4 \\ 3 & -2 \\ -3 & -6 \\ \end{array} \][/tex]
Plotting these points on the graph and drawing the line connecting them.
- Graph [tex]\( y < -3 \)[/tex]:
- Draw a dashed horizontal line at [tex]\( y = -3 \)[/tex] to indicate that [tex]\( y \)[/tex] is strictly less than [tex]\(-3\)[/tex].
- Shade the region below this line.
3. Combine the shaded regions:
- The solution to the system of inequalities is the intersection of the regions shaded for each inequality.
- Thus, we are looking for the region where:
- [tex]\( y \geq \frac{2x}{3} - 4 \)[/tex]
- And [tex]\( y < -3 \)[/tex]
4. Determine the intersection visually or via algebraic approaches:
- We need the [tex]\( y \)[/tex]-values that simultaneously satisfy both conditions.
- Clearly, [tex]\( y \)[/tex] must be less than [tex]\(-3\)[/tex], and it must be on or above [tex]\( \frac{2x}{3} - 4 \)[/tex].
5. Note the boundaries and solution region:
- From plotting, the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex] intersects [tex]\( y < -3 \)[/tex]. We substitute [tex]\( y = -3 \)[/tex] into [tex]\( 2x - 3y = 12 \)[/tex]:
[tex]\[ 2x - 3(-3) = 12 \implies 2x + 9 = 12 \implies 2x = 3 \implies x = \frac{3}{2} \][/tex]
- Thus, at [tex]\( x = \frac{3}{2} \)[/tex] on the line [tex]\( y = -3 \)[/tex], combine this with checking [tex]\( y \)[/tex]-value condition:
- Any point [tex]\( (x, -3) \)[/tex], for [tex]\( x \leq \frac{3}{2} \)[/tex].
### Final Graph Conclusion:
- The final graph shows that the solution region is bounded by:
- The [tex]\( y = \frac{2x}{3} - 4 \)[/tex] line (region above it).
- [tex]\( y = -3 \)[/tex] line (strictly below it).
- And the intersection from [tex]\( x = -\infty \)[/tex] to [tex]\( x = \frac{3}{2} \)[/tex].
Thus, the graph of your solution appears as the region above the line [tex]\( y = \frac{2x}{3} - 4 \)[/tex] but strictly below [tex]\( y = -3 \)[/tex] extending horizontally where conditions intersect.
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