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Choose the graph that represents the following system of inequalities:

[tex]\[
\begin{array}{l}
y \geq -3x + 1 \\
y \leq \frac{1}{2}x + 3
\end{array}
\][/tex]

In each graph, the area for [tex][tex]$f(x)$[/tex][/tex] is shaded and labeled [tex][tex]$A$[/tex][/tex], the area for [tex][tex]$g(x)$[/tex][/tex] is shaded and labeled [tex][tex]$B$[/tex][/tex], and the area where they have shading in common is labeled [tex][tex]$AB$[/tex][/tex].


Sagot :

To solve the system of inequalities and choose the correct graph, we need to graph each inequality on the coordinate plane and find their overlapping region.

Step-by-step solution:

1. Graph the first inequality [tex]\( y \geq -3x + 1 \)[/tex]:

- First, express this inequality as an equation to find the boundary line:
[tex]\[ y = -3x + 1 \][/tex]

- To graph the line, determine the y-intercept and a couple of points:
- The y-intercept is [tex]\( (0, 1) \)[/tex].
- For [tex]\( x = 1 \)[/tex], [tex]\( y = -3(1) + 1 = -2 \)[/tex], so the point is [tex]\( (1, -2) \)[/tex].

- Plot the points [tex]\((0, 1)\)[/tex] and [tex]\((1, -2)\)[/tex], and draw the line through them.

- Since the inequality is [tex]\( y \geq -3x + 1 \)[/tex], shade the region above and including the line.

2. Graph the second inequality [tex]\( y \leq \frac{1}{2}x + 3 \)[/tex]:

- First, express this inequality as an equation to find the boundary line:
[tex]\[ y = \frac{1}{2}x + 3 \][/tex]

- To graph the line, determine the y-intercept and a couple of points:
- The y-intercept is [tex]\( (0, 3) \)[/tex].
- For [tex]\( x = 2 \)[/tex], [tex]\( y = \frac{1}{2}(2) + 3 = 1 + 3 = 4 \)[/tex], so the point is [tex]\( (2, 4) \)[/tex].

- Plot the points [tex]\((0, 3)\)[/tex] and [tex]\((2, 4)\)[/tex], and draw the line through them.

- Since the inequality is [tex]\( y \leq \frac{1}{2}x + 3 \)[/tex], shade the region below and including the line.

3. Determine the overlapping region:

- The overlapping region will be the area where the shaded region from [tex]\( y \geq -3x + 1 \)[/tex] intersects with the shaded region from [tex]\( y \leq \frac{1}{2}x + 3 \)[/tex].

4. Find intersection points (just for the graphing purpose):

- We solve the equations of the boundary lines together:
[tex]\[ -3x + 1 = \frac{1}{2}x + 3 \][/tex]
[tex]\[ -3x - \frac{1}{2}x = 3 - 1 \][/tex]
[tex]\[ -\frac{7}{2}x = 2 \][/tex]
[tex]\[ x = -\frac{2}{\frac{7}{2}} = -\frac{4}{7} \][/tex]
- Substitute [tex]\( x = -\frac{4}{7} \)[/tex] back into one of the equations to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -3\left(-\frac{4}{7}\right) + 1 = \frac{12}{7} + 1 = \frac{12}{7} + \frac{7}{7} = \frac{19}{7} \][/tex]
- The lines intersect at [tex]\( \left(-\frac{4}{7}, \frac{19}{7}\right) \)[/tex].

Final note:
The correct graph will exhibit:
- A shaded region corresponding to [tex]\( y \geq -3x + 1 \)[/tex].
- Another shaded region corresponding to [tex]\( y \leq \frac{1}{2}x + 3 \)[/tex].
- The intersection (overlap) of these shaded areas will be the solution region, labeled as [tex]\( AB \)[/tex].

Use these instructions to find the correct graph among provided options in the question.