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 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 +

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

This problem explores how to find the area enclosed by the x-axis and two lines, x = 3y - 5 and 2y = 7 + 3x, using coordinate geometry and definite integration. Includes a step-by-step breakdown of solving linear equations and calculating area. Suitable for students in Grades 10-12.

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