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To find the area enclosed by the \( x \)-axis and the lines \( x = 3y - 5 \) and \( 2y = 7 + 3x \), let's go through each step to solve the problem.

### Step 1: Rewrite Equations in Slope-Intercept Form
Rewrite each line to express \( y \) as a function of \( x \).

1. **For \( x = 3y - 5 \):**
   \[
   x + 5 = 3y \Rightarrow y = \frac{x + 5}{3}
   \]

2. **For \( 2y = 7 + 3x \):**
   \[
   y = \frac{7 + 3x}{2}
   \]

Now we have the two lines:
- \( y = \frac{x + 5}{3} \)
- \( y = \frac{7 + 3x}{2} \)

### Step 2: Find Points of Intersection
To find the intersection points, set the equations equal to each other:
\[
\frac{x + 5}{3} = \frac{7 + 3x}{2}
\]

Cross-multiplying gives:
\[
2(x + 5) = 3(7 + 3x)
\]
\[
2x + 10 = 21 + 9x
\]
\[
10 - 21 = 9x - 2x
\]
\[
-11 = 7x \Rightarrow x = -\frac{11}{7}
\]

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

Using \( y = \frac{x + 5}{3} \):
\[


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

Learn how to find the area enclosed by the x-axis and the lines x = 3y - 5 and 2y = 7 + 3x. This problem involves converting equations to slope-intercept form, finding intersections, and calculating the area. Suitable for Grades 10-12.

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