 how to calculate the area enclosed by the x-axis and the lines given by x = 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 should follow these steps:

1. **Rewrite the equations in terms of \( y \) as a function of \( x \):**

   - For the line \( x = 3y - 5 \):
     \[
     3y = x + 5
     \]
     \[
     y = \frac{x + 5}{3}
     \]

   - For the line \( 2y = 7 + 3x \):
     \[
     y = \frac{7 + 3x}{2}
     \]

   So, we have two equations:
   \[
   y = \frac{x + 5}{3} \quad \text{and} \quad y = \frac{7 + 3x}{2}.
   \]

2. **Find the points of intersection by setting these two equations equal to each other:**
   \[
   \frac{x + 5}{3} = \frac{7 + 3x}{2}.
   \]

   To eliminate the fractions, cross-multiply:
   \[
   2(x + 5) = 3(7 + 3x).
   \]
   Expanding both sides:
   \[
   2x + 10 = 21 + 9x.
   \]
   Solving for \( x \):
   \[
   -7x = 11 \Rightarrow x = -\frac{11}{7}.
   \]

3. **Substitute \( x = -\frac{11}{7} \) back into one of the original equations to find \( y \):**

   Using \( y = \frac{x + 5}{3} \):
   \[
   y = \frac{-\frac{11}{7} + 5}{3} = \frac{-\frac{11}{7} + \frac{35}{7}}{3} = \frac{\frac{24}{7}}{3} = \frac{24}{21} = \frac{8}{7}.
   \]

   So, the intersection point is \( \left(-\frac{11}{7}, \frac{8}{7}\right) \).

4. **Find the points where each line intersects the \( x \)-axis (where \( y = 0 \)):**

   - For \( y = \frac{x + 5}{3} \):
     \[
     0 = \frac{x + 5}{3} \Rightarrow x = -5.
     \]
     So, this line intersects the \( x \)-axis at \( (-5, 0) \).

   - For \( y = \frac{7 + 3x}{2} \):
     \[
     0 = \frac{7 + 3x}{2} \Rightarrow x = -\frac{7}{3}.
     \]
     So, this line intersects the \( x \)-axis at \( \left(-\frac{7}{3}, 0\right) \).

5. **Set up the integral to calculate the area between the curves from \( x = -5 \) to \( x = -\frac{7}{3} \):**

   The area \( A \) is given by the integral:
   \[
   A = \int_{-5}^{-\frac{7}{3}} \left(\frac{7 + 3x}{2} - \frac{x + 5}{3}\right) \, dx.
   \]

6. **Simplify the integrand and evaluate the integral:**

   Simplify \( \frac{7 + 3x}{2} - \frac{x + 5}{3} \) by finding a common denominator:
   \[
   = \frac{3(7 + 3x) - 2(x + 5)}{6} = \frac{21 + 9x - 2x - 10}{6} = \frac{11 + 7x}{6}.
   \]

   Thus, the integral becomes:
   \[
   A = \int_{-5}^{-\frac{7}{3}} \frac{11 + 7x}{6} \, dx = \frac{1}{6} \int_{-5}^{-\frac{7}{3}} (11 + 7x) \, dx.
   \]

7. **Integrate and find the area:**
   \[
   = \frac{1}{6} \left[ 11x + \frac{7x^2}{2} \right]_{-5}^{-\frac{7}{3}}.
   \]

   Plugging in the limits will yield the enclosed area.

Would you like me to continue with the final calculations or have any questions?

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### Related Questions
1. How do you determine the intersection of two lines?
2. What steps should be followed when setting up an integral for area?
3. How can the intersection with the x-axis help in finding enclosed areas?
4. Why is it necessary to express both equations as \( y \)-functions of \( x \)?
5. What are common approaches to simplify an integrand before calculating?

### Tip
Always double-check the limits of integration to ensure they match the points that enclose the desired area.

How to calculate the area enclosed by the x-axis and the lines given by x = 3y - 5 and 2y = 7 + 3x

Learn to calculate the area enclosed by the x-axis and lines x = 3y - 5, 2y = 7 + 3x with a comprehensive step-by-step solution. This guide covers critical analytic geometry concepts, including integration, finding points of intersection, and setting up integrals. Ideal for students in Grades 11-12.

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