Find the equation of the straight line passing through the point of intersection of the two line `x+2y+3=0 and 3x+4y+7=0` and parallel to the straight line `y-x=8`

Answer

To find the equation of the straight line that passes through the intersection of the two lines \(x + 2y + 3 = 0\) and \(3x + 4y + 7 = 0\), and is parallel to the line \(y - x = 8\), we will follow these steps:

### Step 1: Find the point of intersection of the two lines.

We have the equations:
1. \(x + 2y + 3 = 0\) (Equation 1)
2. \(3x + 4y + 7 = 0\) (Equation 2)

To find the intersection, we can solve these equations simultaneously.

From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = -2y - 3 \]

Now, substitute this expression for \(x\) into Equation 2:
\[ 3(-2y - 3) + 4y + 7 = 0 \]
Expanding this gives:
\[ -6y - 9 + 4y + 7 = 0 \]
Combine like terms:
\[ -2y - 2 = 0 \]
Solving for \(y\):
\[ -2y = 2 \implies y = -1 \]

Now substitute \(y = -1\) back into the expression for \(x\):
\[ x = -2(-1) - 3 = 2 - 3 = -1 \]

Thus, the point of intersection is \((-1, -1)\).

### Step 2: Determine the slope of the line parallel to \(y - x = 8\).

Rearranging the equation \(y - x = 8\) into slope-intercept form \(y = mx + b\):
\[ y = x + 8 \]
The slope \(m\) of this line is \(1\).

### Step 3: Use the point-slope form of the line equation.

Since we need a line that is parallel to the given line and passes through the point \((-1, -1)\), we can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = 1\), \(x_1 = -1\), and \(y_1 = -1\):
\[ y - (-1) = 1(x - (-1)) \]
This simplifies to:
\[ y + 1 = x + 1 \]
Rearranging gives:
\[ y = x \]

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