Find the equation of the line passing through the intersection of the lines `2x+y=5\ a n d\ x+3y+8=0` and parallel to the line `3x+4y=7.`

Explanation

To find the equation of the line passing through the intersection of the lines \(2x + y = 5\) and \(x + 3y + 8 = 0\), and parallel to the line \(3x + 4y = 7\), we can follow these steps:

### Step 1: Identify the given lines
We have two lines:
1. \(L_1: 2x + y - 5 = 0\)
2. \(L_2: x + 3y + 8 = 0\)

And we need to find a line that is parallel to:
3. \(L_3: 3x + 4y - 7 = 0\)

### Step 2: Find the intersection of the two lines
To find the intersection of \(L_1\) and \(L_2\), we can solve the equations simultaneously.

From \(L_1\):
\[ y = 5 - 2x \quad (1) \]

Substituting equation (1) into \(L_2\):
\[ x + 3(5 - 2x) + 8 = 0 \]
\[ x + 15 - 6x + 8 = 0 \]
\[ -5x + 23 = 0 \]
\[ 5x = 23 \implies x = \frac{23}{5} \]

Now substitute \(x\) back into equation (1) to find \(y\):
\[ y = 5 - 2\left(\frac{23}{5}\right) = 5 - \frac{46}{5} = \frac{25 - 46}{5} = \frac{-21}{5} \]

So, the intersection point is:
\[ \left(\frac{23}{5}, \frac{-21}{5}\right) \]

### Step 3: Form the equation of the line through the intersection
Any line passing through the intersection of \(L_1\) and \(L_2\) can be expressed as:
\[ L = L_1 + \lambda L_2 = 0 \]
Substituting \(L_1\) and \(L_2\):
\[ (2x + y - 5) + \lambda (x + 3y + 8) = 0 \]
Expanding this:
\[ 2x + y - 5 + \lambda x + 3\lambda y + 8\lambda = 0 \]
Combining like terms:
\[ (2 + \lambda)x + (1 + 3\lambda)y + (8\lambda - 5) = 0 \]

### Step 4: Find the condition for parallelism
For the line to be parallel to \(L_3: 3x + 4y - 7 = 0\), the ratios of the coefficients must be equal:
\[ \frac{2 + \lambda}{3} = \frac{1 + 3\lambda}{4} \]

Cross-multiplying gives:
\[ 4(2 + \lambda) = 3(1 + 3\lambda) \]
Expanding both sides:
\[ 8 + 4\lambda = 3 + 9\lambda \]
Rearranging gives:
\[ 8 - 3 = 9\lambda - 4\lambda \]
\[ 5 = 5\lambda \implies \lambda = 1 \]

### Step 5: Substitute \(\lambda\) back into the line equation
Substituting \(\lambda = 1\) into the line equation:
\[ (2 + 1)x + (1 + 3 \cdot 1)y + (8 \cdot 1 - 5) = 0 \]
This simplifies to:
\[ 3x + 4y + 3 = 0 \]

### Final Answer
The equation of the required line is:
\[ 3x + 4y + 3 = 0 \]