Find the equation of the line through the intersection of lines `3x+4y=7` and `x- y+2=0`and whose slope is 5.

Explanation

To find the equation of the line that passes through the intersection of the lines \(3x + 4y = 7\) and \(x - y + 2 = 0\) with a slope of 5, we can follow these steps:

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

We will solve the equations \(3x + 4y = 7\) and \(x - y + 2 = 0\) simultaneously.

1. From the second equation, express \(x\) in terms of \(y\):
\[ x = y - 2 \]

2. Substitute \(x\) in the first equation:
\[ 3(y - 2) + 4y = 7 \]
\[ 3y - 6 + 4y = 7 \]
\[ 7y - 6 = 7 \]
\[ 7y = 13 \]
\[ y = \frac{13}{7} \]

3. Now substitute \(y\) back to find \(x\):
\[ x = \frac{13}{7} - 2 = \frac{13}{7} - \frac{14}{7} = -\frac{1}{7} \]

So, the intersection point is \(\left(-\frac{1}{7}, \frac{13}{7}\right)\).

### Step 2: Use the point-slope form of the equation of a line

The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is the point through which the line passes and \(m\) is the slope.

Here, \((x_1, y_1) = \left(-\frac{1}{7}, \frac{13}{7}\right)\) and \(m = 5\).

Substituting these values into the point-slope form:
\[ y - \frac{13}{7} = 5\left(x + \frac{1}{7}\right) \]

### Step 3: Simplify the equation

1. Distribute the slope on the right side:
\[ y - \frac{13}{7} = 5x + \frac{5}{7} \]

2. Add \(\frac{13}{7}\) to both sides:
\[ y = 5x + \frac{5}{7} + \frac{13}{7} \]
\[ y = 5x + \frac{18}{7} \]

### Step 4: Convert to standard form

To convert \(y = 5x + \frac{18}{7}\) into standard form \(Ax + By + C = 0\):

1. Rearranging gives:
\[ -5x + y - \frac{18}{7} = 0 \]

2. Multiply through by 7 to eliminate the fraction:
\[ -35x + 7y - 18 = 0 \]

3. Rearranging gives:
\[ 35x - 7y + 18 = 0 \]

Thus, the required equation of the line is:
\[ 35x - 7y + 18 = 0 \]