To find the area enclosed by the x-axis and the straight lines $x = 3y - 5$ and $2y = 7 + 3x$, we need to follow a systematic approach.

Step 1: Write equations of the lines in terms of $y$

Given the two equations:

1. $x = 3y - 5$
2. $2y = 7 + 3x$

Let's rewrite them both in the standard form $y = f(x)$.

Equation 1: $x = 3y - 5$

Solve for $y$: $y = \frac{x + 5}{3}$

Equation 2: $2y = 7 + 3x$

Solve for $y$: $y = \frac{7 + 3x}{2}$

Now, the two lines are: $y = \frac{x + 5}{3} \quad \text{(Line 1)}$ $y = \frac{7 + 3x}{2} \quad \text{(Line 2)}$

Step 2: Find points of intersection

To find the points of intersection with each other and with the x-axis, we first set the two equations equal to each other to find their intersection:

$\frac{x + 5}{3} = \frac{7 + 3x}{2}$

Multiply through by 6 (the least common denominator):

$2(x + 5) = 3(7 + 3x)$

Expand both sides:

$2x + 10 = 21 + 9x$

Simplify:

$10 - 21 = 9x - 2x$ $-11 = 7x$ $x = -\frac{11}{7}$

Now substitute $x = -\frac{11}{7}$ into either equation to find $y$. Using the first equation:

$y = \frac{x + 5}{3} = \frac{-\frac{11}{7} + 5}{3}$ First, convert 5 to a fraction with denominator 7:

$5 = \frac{35}{7}$ Now:

$y = \frac{-\frac{11}{7} + \frac{35}{7}}{3} = \frac{\frac{24}{7}}{3} = \frac{24}{21} = \frac{8}{7}$

So, the lines intersect at $\left(-\frac{11}{7}, \frac{8}{7}\right)$.

Step 3: Find intercepts with the x-axis

The intercepts with the x-axis occur where $y = 0$.

For Line 1: $y = \frac{x + 5}{3}$

Set $y = 0$:

$0 = \frac{x + 5}{3}$ $x = -5$

So, Line 1 intersects the x-axis at $x = -5$.

For Line 2: $y = \frac{7 + 3x}{2}$

Set $y = 0$:

$0 = \frac{7 + 3x}{2}$ $7 + 3x = 0$ $x = -\frac{7}{3}$

So, Line 2 intersects the x-axis at $x = -\frac{7}{3}$.

Step 4: Set up the integral to find the area

The area enclosed by the x-axis and the lines can be found by integrating the difference between the two functions from the leftmost x-intercept to the rightmost x-intercept.

The limits of integration are from $x = -5$ to $x = -\frac{7}{3}$, and the functions are:

$y_1 = \frac{x + 5}{3}, \quad y_2 = \frac{7 + 3x}{2}$

The enclosed area is given by:

$\text{Area} = \int_{-5}^{-\frac{7}{3}} \left( \frac{7 + 3x}{2} - \frac{x + 5}{3} \right) dx$

Step 5: Compute the integral

First, simplify the integrand:

[ \frac{7 + 3x}{2} - \frac{x + 5}{3} = \frac{3(7 + 3x) - 2(x +