Question
Solve the system of inequalities in two variables
3x−2y≤−2
x+2y≥6
3x−2y≤−2
x+2y≥6
Answers
GPT-4o mini
To solve the system of inequalities:
1. **Graph the inequalities**:
- For the inequality \(3x - 2y \leq -2\):
- Rewrite in slope-intercept form:
\[
-2y \leq -3x - 2 \implies y \geq \frac{3}{2}x + 1
\]
- The line \(y = \frac{3}{2}x + 1\) is drawn with a dashed line (since it is \(\leq\)) and shaded above it.
- For the inequality \(x + 2y \geq 6\):
- Rewrite in slope-intercept form:
\[
2y \geq -x + 6 \implies y \geq -\frac{1}{2}x + 3
\]
- The line \(y = -\frac{1}{2}x + 3\) is drawn with a dashed line (since it is \(\geq\)) and shaded above it.
2. **Find the intersection points** of the boundaries of the inequalities:
- Set the equations equal to find the intersection:
\[
\frac{3}{2}x + 1 = -\frac{1}{2}x + 3
\]
To eliminate fractions, multiply through by 2:
\[
3x + 2 = -x + 6
\]
Rearranging gives:
\[
4x = 4 \implies x = 1
\]
Substitute \(x = 1\) back into one of the original line equations:
\[
y = \frac{3}{2}(1) + 1 = \frac{5}{2}
\]
- Thus, the intersection point is \((1, \frac{5}{2})\).
3. **Identify the feasible region**:
- The feasible region is where the shaded areas for both inequalities overlap.
- Since the graphs are linear, the region will be a polygon or part of a plane bounded by the lines.
4. **Test a point** inside the feasible region:
- For example, take the point \((0, 0)\):
- Check \(3(0) - 2(0) \leq -2\), which is false.
- Check \(0 + 2(0) \geq 6\), which is also false.
- Therefore, the origin is not in the feasible region. You might try another point, like \((0, 4)\):
- Check \(3(0) - 2(4) \leq -2\), which gives \(-8 \leq -2\) (true).
- Check \(0 + 2(4) \geq 6\), which gives \(8 \geq 6\) (true).
- Hence, \((0, 4)\) is in the feasible region.
5. **Conclusion**:
- The solution to the system of inequalities consists of all the points \((x, y)\) that lie in the overlapping shaded region created by the two inequalities. This can be expressed as a set of ordered pairs or a graphical representation. The lines \(y = \frac{3}{2}x + 1\) and \(y = -\frac{1}{2}x + 3\) serve as boundaries for this feasible region.
1. **Graph the inequalities**:
- For the inequality \(3x - 2y \leq -2\):
- Rewrite in slope-intercept form:
\[
-2y \leq -3x - 2 \implies y \geq \frac{3}{2}x + 1
\]
- The line \(y = \frac{3}{2}x + 1\) is drawn with a dashed line (since it is \(\leq\)) and shaded above it.
- For the inequality \(x + 2y \geq 6\):
- Rewrite in slope-intercept form:
\[
2y \geq -x + 6 \implies y \geq -\frac{1}{2}x + 3
\]
- The line \(y = -\frac{1}{2}x + 3\) is drawn with a dashed line (since it is \(\geq\)) and shaded above it.
2. **Find the intersection points** of the boundaries of the inequalities:
- Set the equations equal to find the intersection:
\[
\frac{3}{2}x + 1 = -\frac{1}{2}x + 3
\]
To eliminate fractions, multiply through by 2:
\[
3x + 2 = -x + 6
\]
Rearranging gives:
\[
4x = 4 \implies x = 1
\]
Substitute \(x = 1\) back into one of the original line equations:
\[
y = \frac{3}{2}(1) + 1 = \frac{5}{2}
\]
- Thus, the intersection point is \((1, \frac{5}{2})\).
3. **Identify the feasible region**:
- The feasible region is where the shaded areas for both inequalities overlap.
- Since the graphs are linear, the region will be a polygon or part of a plane bounded by the lines.
4. **Test a point** inside the feasible region:
- For example, take the point \((0, 0)\):
- Check \(3(0) - 2(0) \leq -2\), which is false.
- Check \(0 + 2(0) \geq 6\), which is also false.
- Therefore, the origin is not in the feasible region. You might try another point, like \((0, 4)\):
- Check \(3(0) - 2(4) \leq -2\), which gives \(-8 \leq -2\) (true).
- Check \(0 + 2(4) \geq 6\), which gives \(8 \geq 6\) (true).
- Hence, \((0, 4)\) is in the feasible region.
5. **Conclusion**:
- The solution to the system of inequalities consists of all the points \((x, y)\) that lie in the overlapping shaded region created by the two inequalities. This can be expressed as a set of ordered pairs or a graphical representation. The lines \(y = \frac{3}{2}x + 1\) and \(y = -\frac{1}{2}x + 3\) serve as boundaries for this feasible region.