Find the area enclosed by the x-axis and the straight lines 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 can proceed as follows:

### Step 1: Rewrite the Equations in Slope-Intercept Form
1. **For \(x = 3y - 5\):**
   \[
   3y = x + 5 \Rightarrow y = \frac{x + 5}{3} \Rightarrow y = \frac{1}{3}x + \frac{5}{3}
   \]
   So the equation of the first line is \(y = \frac{1}{3}x + \frac{5}{3}\).

2. **For \(2y = 7 + 3x\):**
   \[
   y = \frac{7 + 3x}{2} \Rightarrow y = \frac{3}{2}x + \frac{7}{2}
   \]
   So the equation of the second line is \(y = \frac{3}{2}x + \frac{7}{2}\).

### Step 2: Find the Points of Intersection
1. **Intersection of the Lines:**
   Set the two expressions for \(y\) equal to each other:
   \[
   \frac{1}{3}x + \frac{5}{3} = \frac{3}{2}x + \frac{7}{2}
   \]
   Multiply through by 6 to clear the fractions:
   \[
   2x + 10 = 9x + 21
   \]
   Rearranging terms:
   \[
   -7x = 11 \Rightarrow x = -\frac{11}{7}
   \]
   Substitute \(x = -\frac{11}{7}\) into either equation to find \(y\). Using \(y = \frac{1}{3}x + \frac{5}{3}\):
   \[
   y = \frac{1}{3} \left(-\frac{11}{7}\right) + \frac{5}{3} = -\frac{11}{21} + \frac{35}{21} = \frac{24}{21} = \frac{8}{7}
   \]
   So, the lines intersect at \( \left(-\frac{11}{7}, \frac{8}{7}\right) \).

2. **Intersection with the \(x\)-axis:**
   To find the points where each line intersects the \(x\)-axis, set \(y = 0\) in each equation:
   
   - For \(y = \frac{1}{3}x + \frac{5}{3}\):
     \[
     0 = \frac{1}{3}x + \frac{5}{3} \Rightarrow x = -5
     \]
     So, one intercept is \((-5, 0)\).

   - For \(y = \frac{3}{2}x + \frac{7}{2}\):
     \[
     0 = \frac{3}{2}x + \frac{7}{2} \Rightarrow x = -\frac{7}{3}
     \]
     So, the other intercept is \(\left(-\frac{7}{3}, 0\right)\).

### Step 3: Set Up the Integral for the Area
The area enclosed by the lines and the \(x\)-axis is the area between \((-5, 0)\), \(\left(-\frac{7}{3}, 0\right)\), and the intersection point \(\left(-\frac{11}{7}, \frac{8}{7}\right)\). We will integrate the difference between the two lines over the interval \([-5, -\frac{7}{3}]\).

1. **Integral Setup:**
   \[
   \text{Area} = \int_{-5}^{-\frac{7}{3}} \left(\frac{3}{2}x + \frac{7}{2} - \left(\frac{1}{3}x + \frac{5}{3}\right)\right) \, dx
   \]

2. **Simplify the Integrand:**
   \[
   = \int_{-5}^{-\frac{7}{3}} \left(\frac{3}{2}x - \frac{1}{3}x + \frac{7}{2} - \frac{5}{3}\right) \, dx
   \]
   Combine like terms:
   \[
   = \int_{-5}^{-\frac{7}{3}} \left(\frac{9x - 2x}{6} + \frac{21 - 10}{6}\right) \, dx
   \]
   Simplify

Learn how to calculate the area enclosed by the x-axis and two lines, x=3y-5 and 2y=7+3x, using integration and analytic geometry techniques. This detailed solution is ideal for students in Grades 11-12 and covers linear equations and the application of the Fundamental Theorem of Calculus.

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