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Sagot :
To solve and graph the system of inequalities, follow these steps:
### Step 1: Rewrite the inequalities in slopeintercept form (i.e., [tex]\( y = mx + b \)[/tex]).
1. First Inequality: [tex]\( 3y > 2x + 12 \)[/tex]
Divide all terms by 3 to isolate [tex]\( y \)[/tex]:
[tex]\[ y > \frac{2}{3}x + 4 \][/tex]
2. Second Inequality: [tex]\( 2x + y \leq 5 \)[/tex]
Subtract [tex]\( 2x \)[/tex] from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y \leq 2x  5 \][/tex]
### Step 2: Graph the boundary lines for each inequality as if they were equalities.
1. Boundary Line for [tex]\( y = \frac{2}{3}x + 4 \)[/tex]
 This line has a slope ([tex]\( m \)[/tex]) of [tex]\( \frac{2}{3} \)[/tex] and a yintercept ([tex]\( b \)[/tex]) of 4.
 Plot the yintercept at (0, 4).
 Use the slope to find another point: from (0, 4), move up 2 units and right 3 units to (3, 6).
 Draw a dashed line through these points because the inequality is [tex]\( > \)[/tex], not [tex]\( \geq \)[/tex].
2. Boundary Line for [tex]\( y = 2x  5 \)[/tex]
 This line has a slope ([tex]\( m \)[/tex]) of [tex]\( 2 \)[/tex] and a yintercept ([tex]\( b \)[/tex]) of 5.
 Plot the yintercept at (0, 5).
 Use the slope to find another point: from (0, 5), move down 2 units and right 1 unit to (1, 7).
 Draw a solid line through these points because the inequality is [tex]\( \leq \)[/tex], which includes the boundary.
### Step 3: Determine which side of each boundary line represents the solution to the inequality.
1. For [tex]\( y > \frac{2}{3}x + 4 \)[/tex]
 Choose a test point not on the line, such as (0, 0).
 Substitute into the inequality: [tex]\( 0 \stackrel{?}{>} \frac{2}{3}(0) + 4 \rightarrow 0 \stackrel{?}{>} 4 \)[/tex].
 This is false, so the shading is above the line where the inequality holds.
2. For [tex]\( y \leq 2x  5 \)[/tex]
 Choose a test point not on the line, such as (0, 0).
 Substitute into the inequality: [tex]\( 0 \stackrel{?}{\leq} 2(0)  5 \rightarrow 0 \stackrel{?}{\leq} 5 \)[/tex].
 This is false, so the shading is below the line where the inequality holds.
### Step 4: Identify the overlapping region.
 The region that satisfies both inequalities is where the shaded areas overlap.
 For clarity:
 Shade above the dashed line [tex]\( y = \frac{2}{3}x + 4 \)[/tex].
 Shade below the solid line [tex]\( y = 2x  5 \)[/tex].
### Step 5: Graph the solution on the coordinate plane.
1. Dashed Line ( [tex]\( y = \frac{2}{3}x + 4 \)[/tex] )
 Starts at (0, 4), passes through (3, 6), and extends in both directions with a dashed pattern.
 Shade above this line.
2. Solid Line ( [tex]\( y = 2x  5 \)[/tex] )
 Starts at (0, 5), passes through (1, 7), and extends in both directions with a solid pattern.
 Shade below this line.
3. Overlapping region
 The area where the shadings from both inequalities overlap is the solution region.
Here's a graph illustrating the solution:
```plaintext
y

15 
10 
5  _____________
0 __________0__________
5 _________ __________ ____
10 
15 
10 5 0 5 10 x
```
(Note: Graphing tools cannot be rendered here directly, so you need to manually draw this on graph paper or using a graphing tool.)
### Step 1: Rewrite the inequalities in slopeintercept form (i.e., [tex]\( y = mx + b \)[/tex]).
1. First Inequality: [tex]\( 3y > 2x + 12 \)[/tex]
Divide all terms by 3 to isolate [tex]\( y \)[/tex]:
[tex]\[ y > \frac{2}{3}x + 4 \][/tex]
2. Second Inequality: [tex]\( 2x + y \leq 5 \)[/tex]
Subtract [tex]\( 2x \)[/tex] from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y \leq 2x  5 \][/tex]
### Step 2: Graph the boundary lines for each inequality as if they were equalities.
1. Boundary Line for [tex]\( y = \frac{2}{3}x + 4 \)[/tex]
 This line has a slope ([tex]\( m \)[/tex]) of [tex]\( \frac{2}{3} \)[/tex] and a yintercept ([tex]\( b \)[/tex]) of 4.
 Plot the yintercept at (0, 4).
 Use the slope to find another point: from (0, 4), move up 2 units and right 3 units to (3, 6).
 Draw a dashed line through these points because the inequality is [tex]\( > \)[/tex], not [tex]\( \geq \)[/tex].
2. Boundary Line for [tex]\( y = 2x  5 \)[/tex]
 This line has a slope ([tex]\( m \)[/tex]) of [tex]\( 2 \)[/tex] and a yintercept ([tex]\( b \)[/tex]) of 5.
 Plot the yintercept at (0, 5).
 Use the slope to find another point: from (0, 5), move down 2 units and right 1 unit to (1, 7).
 Draw a solid line through these points because the inequality is [tex]\( \leq \)[/tex], which includes the boundary.
### Step 3: Determine which side of each boundary line represents the solution to the inequality.
1. For [tex]\( y > \frac{2}{3}x + 4 \)[/tex]
 Choose a test point not on the line, such as (0, 0).
 Substitute into the inequality: [tex]\( 0 \stackrel{?}{>} \frac{2}{3}(0) + 4 \rightarrow 0 \stackrel{?}{>} 4 \)[/tex].
 This is false, so the shading is above the line where the inequality holds.
2. For [tex]\( y \leq 2x  5 \)[/tex]
 Choose a test point not on the line, such as (0, 0).
 Substitute into the inequality: [tex]\( 0 \stackrel{?}{\leq} 2(0)  5 \rightarrow 0 \stackrel{?}{\leq} 5 \)[/tex].
 This is false, so the shading is below the line where the inequality holds.
### Step 4: Identify the overlapping region.
 The region that satisfies both inequalities is where the shaded areas overlap.
 For clarity:
 Shade above the dashed line [tex]\( y = \frac{2}{3}x + 4 \)[/tex].
 Shade below the solid line [tex]\( y = 2x  5 \)[/tex].
### Step 5: Graph the solution on the coordinate plane.
1. Dashed Line ( [tex]\( y = \frac{2}{3}x + 4 \)[/tex] )
 Starts at (0, 4), passes through (3, 6), and extends in both directions with a dashed pattern.
 Shade above this line.
2. Solid Line ( [tex]\( y = 2x  5 \)[/tex] )
 Starts at (0, 5), passes through (1, 7), and extends in both directions with a solid pattern.
 Shade below this line.
3. Overlapping region
 The area where the shadings from both inequalities overlap is the solution region.
Here's a graph illustrating the solution:
```plaintext
y

15 
10 
5  _____________
0 __________0__________
5 _________ __________ ____
10 
15 
10 5 0 5 10 x
```
(Note: Graphing tools cannot be rendered here directly, so you need to manually draw this on graph paper or using a graphing tool.)
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