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To determine which graph represents the solution to the system of inequalities:
[tex]\[ \begin{array}{l} x + y < 4 \\ 2x - 3y \geq 12 \end{array} \][/tex]
we need to understand and plot each inequality separately and then find the region where both conditions are satisfied.
### Step 1: Plotting [tex]\( x + y < 4 \)[/tex]
1. Rewrite the inequality as an equation to plot the boundary line:
[tex]\[ x + y = 4 \][/tex]
This is a straight line. To find points on this line, we can choose two points:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 4 \)[/tex]. Hence, the point is (0, 4).
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 4 \)[/tex]. Hence, the point is (4, 0).
2. Draw the line passing through these two points (0,4) and (4,0).
3. Since the inequality is [tex]\( x + y < 4 \)[/tex], shade the region below the line because this inequality represents all points where the sum of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is less than 4.
### Step 2: Plotting [tex]\( 2x - 3y \geq 12 \)[/tex]
1. Rewrite the inequality as an equation to plot the boundary line:
[tex]\[ 2x - 3y = 12 \][/tex]
This is a straight line. To find points on this line, we can choose:
- When [tex]\( x = 0 \)[/tex], solving for [tex]\( y \)[/tex]: [tex]\( -3y = 12 \)[/tex] gives [tex]\( y = -4 \)[/tex]. So, the point is (0, -4).
- When [tex]\( y = 0 \)[/tex], solving for [tex]\( x \)[/tex]: [tex]\( 2x = 12 \)[/tex] gives [tex]\( x = 6 \)[/tex]. So, the point is (6, 0).
2. Draw the line passing through these two points (0, -4) and (6, 0).
3. Since the inequality is [tex]\( 2x - 3y \geq 12 \)[/tex], shade the region above the line because this inequality represents all points where [tex]\( 2x - 3y \)[/tex] is greater than or equal to 12.
### Step 3: Finding the solution region
The solution to the system of inequalities is the region where both shaded areas overlap. Graphically:
- The line [tex]\( x + y = 4 \)[/tex] bounds the region below it for [tex]\( x + y < 4 \)[/tex].
- The line [tex]\( 2x - 3y = 12 \)[/tex] bounds the region above it for [tex]\( 2x - 3y \geq 12 \)[/tex].
Find the intersection of the two boundaries:
Set [tex]\( x + y = 4 \)[/tex] and [tex]\( 2x - 3y = 12 \)[/tex]. Solve these equations simultaneously as a system of linear equations:
1. From [tex]\( x + y = 4 \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y = 4 - x \][/tex]
2. Substitute [tex]\( y = 4 - x \)[/tex] into [tex]\( 2x - 3y = 12 \)[/tex]:
[tex]\[ 2x - 3(4 - x) = 12 \\ 2x - 12 + 3x = 12 \\ 5x - 12 = 12 \\ 5x = 24 \\ x = \frac{24}{5} = 4.8 \][/tex]
3. Substitute [tex]\( x = 4.8 \)[/tex] back into [tex]\( y = 4 - x \)[/tex]:
[tex]\[ y = 4 - 4.8 = -0.8 \][/tex]
The intersection point is [tex]\( (4.8, -0.8) \)[/tex].
### Conclusion
The region where both shading overlap is bounded by the lines with the intersection at [tex]\( (4.8, -0.8) \)[/tex]:
- Below [tex]\( x + y < 4 \)[/tex].
- Above [tex]\( 2x - 3y \geq 12 \)[/tex].
Based on the described constraints and without visual aids, the correct graph should show a shaded region below the line [tex]\( x + y = 4 \)[/tex] and above the line [tex]\( 2x - 3y = 12 \)[/tex], with intersection at the point [tex]\( (4.8, -0.8) \)[/tex].
Thus, you should choose the graph that represents this overlapping region as the solution. Without actual graph options provided as stated "A. [tex]$\qquad$[/tex] B. [tex]\(y \mp 10\)[/tex]," which seems incomplete, we can't specify which graphic is the accurate match. Thus, please verify against the available graphs accordingly.
[tex]\[ \begin{array}{l} x + y < 4 \\ 2x - 3y \geq 12 \end{array} \][/tex]
we need to understand and plot each inequality separately and then find the region where both conditions are satisfied.
### Step 1: Plotting [tex]\( x + y < 4 \)[/tex]
1. Rewrite the inequality as an equation to plot the boundary line:
[tex]\[ x + y = 4 \][/tex]
This is a straight line. To find points on this line, we can choose two points:
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 4 \)[/tex]. Hence, the point is (0, 4).
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 4 \)[/tex]. Hence, the point is (4, 0).
2. Draw the line passing through these two points (0,4) and (4,0).
3. Since the inequality is [tex]\( x + y < 4 \)[/tex], shade the region below the line because this inequality represents all points where the sum of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is less than 4.
### Step 2: Plotting [tex]\( 2x - 3y \geq 12 \)[/tex]
1. Rewrite the inequality as an equation to plot the boundary line:
[tex]\[ 2x - 3y = 12 \][/tex]
This is a straight line. To find points on this line, we can choose:
- When [tex]\( x = 0 \)[/tex], solving for [tex]\( y \)[/tex]: [tex]\( -3y = 12 \)[/tex] gives [tex]\( y = -4 \)[/tex]. So, the point is (0, -4).
- When [tex]\( y = 0 \)[/tex], solving for [tex]\( x \)[/tex]: [tex]\( 2x = 12 \)[/tex] gives [tex]\( x = 6 \)[/tex]. So, the point is (6, 0).
2. Draw the line passing through these two points (0, -4) and (6, 0).
3. Since the inequality is [tex]\( 2x - 3y \geq 12 \)[/tex], shade the region above the line because this inequality represents all points where [tex]\( 2x - 3y \)[/tex] is greater than or equal to 12.
### Step 3: Finding the solution region
The solution to the system of inequalities is the region where both shaded areas overlap. Graphically:
- The line [tex]\( x + y = 4 \)[/tex] bounds the region below it for [tex]\( x + y < 4 \)[/tex].
- The line [tex]\( 2x - 3y = 12 \)[/tex] bounds the region above it for [tex]\( 2x - 3y \geq 12 \)[/tex].
Find the intersection of the two boundaries:
Set [tex]\( x + y = 4 \)[/tex] and [tex]\( 2x - 3y = 12 \)[/tex]. Solve these equations simultaneously as a system of linear equations:
1. From [tex]\( x + y = 4 \)[/tex], solve for [tex]\( y \)[/tex]:
[tex]\[ y = 4 - x \][/tex]
2. Substitute [tex]\( y = 4 - x \)[/tex] into [tex]\( 2x - 3y = 12 \)[/tex]:
[tex]\[ 2x - 3(4 - x) = 12 \\ 2x - 12 + 3x = 12 \\ 5x - 12 = 12 \\ 5x = 24 \\ x = \frac{24}{5} = 4.8 \][/tex]
3. Substitute [tex]\( x = 4.8 \)[/tex] back into [tex]\( y = 4 - x \)[/tex]:
[tex]\[ y = 4 - 4.8 = -0.8 \][/tex]
The intersection point is [tex]\( (4.8, -0.8) \)[/tex].
### Conclusion
The region where both shading overlap is bounded by the lines with the intersection at [tex]\( (4.8, -0.8) \)[/tex]:
- Below [tex]\( x + y < 4 \)[/tex].
- Above [tex]\( 2x - 3y \geq 12 \)[/tex].
Based on the described constraints and without visual aids, the correct graph should show a shaded region below the line [tex]\( x + y = 4 \)[/tex] and above the line [tex]\( 2x - 3y = 12 \)[/tex], with intersection at the point [tex]\( (4.8, -0.8) \)[/tex].
Thus, you should choose the graph that represents this overlapping region as the solution. Without actual graph options provided as stated "A. [tex]$\qquad$[/tex] B. [tex]\(y \mp 10\)[/tex]," which seems incomplete, we can't specify which graphic is the accurate match. Thus, please verify against the available graphs accordingly.
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