Find the area enclosed by the x-axis and the straight linesx=3y-5 and 2y=7+3x

To find the area enclosed by the \( x \)-axis and the lines \( x = 3y - 5 \) and \( 2y = 7 + 3x \), we will:

1. **Rewrite the equations in slope-intercept form:**
   - Given:
     \[
     x = 3y - 5
     \]
     Rewrite it as:
     \[
     y = \frac{x + 5}{3}
     \]

   - For the second equation:
     \[
     2y = 7 + 3x
     \]
     Divide by 2:
     \[
     y = \frac{7}{2} + \frac{3}{2}x
     \]

2. **Find the points of intersection:**
   - Solve \( \frac{x + 5}{3} = \frac{7}{2} + \frac{3}{2}x \):
     \[
     \frac{x + 5}{3} = \frac{7}{2} + \frac{3}{2}x
     \]
     Cross-multiply to clear the fractions:
     \[
     2(x + 5) = 3(7 + 3x)
     \]
     Expand:
     \[
     2x + 10 = 21 + 9x
     \]
     Rearrange to solve for \( x \):
     \[
     10 - 21 = 9x - 2x
     \]
     \[
     -11 = 7x \implies x = -\frac{11}{7}
     \]

   - Substitute \( x = -\frac{11}{7} \) into \( y = \frac{x + 5}{3} \):
     \[
     y = \frac{-\frac{11}{7} + 5}{3} = \frac{-\frac{11}{7} + \frac{35}{7}}{3} = \frac{\frac{24}{7}}{3} = \frac{24}{21} = \frac{8}{7}
     \]
   Thus, the intersection point is \( \left(-\frac{11}{7}, \frac{8}{7}\right) \).

3. **Find the x-intercepts of both lines:**
   - For \( y = \frac{x + 5}{3} \), set \( y = 0 \):
     \[
     0 = \frac{x + 5}{3} \implies x = -5
     \]
   - For \( y = \frac{7}{2} + \frac{3}{2}x \), set \( y = 0 \):
     \[
     0 = \frac{7}{2} + \frac{3}{2}x \implies \frac{3}{2}x = -\frac{7}{2}
     \]
     \[
     x = -\frac{7}{3}
     \]

4. **Calculate the area enclosed using the vertices:**
   The vertices of the enclosed region are:
   \[
   (-5, 0), \left(-\frac{7}{3}, 0\right), \left(-\frac{11}{7}, \frac{8}{7}\right)
   \]

   Use the Shoelace Theorem (area formula for a polygon):
   \[
   \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_1 - (y_1x_2 + y_2x_3 + y_3x_1) \right|
   \]
   Substitute \( (x_1, y_1) = (-5, 0), (x_2, y_2) = \left(-\frac{7}{3}, 0\right), (x_3, y_3) = \left(-\frac{11}{7}, \frac{8}{7}\right) \):
   \[
   \text{Area} = \frac{1}{2} \left| (-5)(0) + \left(-\frac{7}{3}\right)\left(\frac{8}{7}\right) + \left(-\frac{11}{7}\right)(0) - (0 \cdot \left(-\frac{7}{3}\right) + 0 \cdot \left(-\frac{11}{7}\right) + \frac{8}{7} \cdot (-5)) \right|
   \]
   \[
   = \frac{1}{2} \left| 0 + \left(-\frac{56}{21}\right) + 0 - (0 + 0 - \frac{40}{7}) \right|
   \]
   \[
   = \frac{1}{2} \left| -\frac{56}{21} + \frac{120}{21} \right|
   \]
   \[
   = \frac{1}{2} \left| \frac{64}{21} \right|
   \]
   \[
   = \frac{32}{21}
   \]

Thus, the area enclosed by the \( x \)-axis and the two lines is \( \frac{32}{21} \) square units.

Would you like further details or have any questions?

### Related Questions:
1. How do you find the intersection points of two lines algebraically?
2. What is the Shoelace Theorem, and how does it help in finding areas of polygons?
3. How do we convert different forms of linear equations into slope-intercept form?
4. Why is it important to use absolute values when calculating the area of geometric figures?
5. How does the arrangement of points affect the calculation of polygonal areas?

### Tip:
When finding the area enclosed by curves or lines, plotting the graph can provide visual insights into the region being calculated.

Find the area enclosed by the x-axis and the straight lines x=3y-5 and 2y=7+3x

Learn how to calculate the area enclosed by the x-axis and the lines x=3y-5 and 2y=7+3x. This problem explores concepts from algebra and geometry, including slope-intercept form and the Shoelace Theorem, providing a step-by-step solution. Suitable for students in Grades 10-12.

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