Find the equation of the line through the intersection of `x-y=1 and 2x-3y+1=0` and parallel to `3x+4y=12`.

Answer

To find the equation of the line through the intersection of the lines \( x - y = 1 \) and \( 2x - 3y + 1 = 0 \), and parallel to the line \( 3x + 4y = 12 \), we can follow these steps:

### Step 1: Find the intersection point of the two lines
We have the equations:
1. \( x - y = 1 \) (Equation 1)
2. \( 2x - 3y + 1 = 0 \) (Equation 2)

First, we can rearrange Equation 1 to express \( x \) in terms of \( y \):
\[ x = y + 1 \]

Now, substitute \( x \) in Equation 2:
\[ 2(y + 1) - 3y + 1 = 0 \]
Expanding this gives:
\[ 2y + 2 - 3y + 1 = 0 \]
Combining like terms:
\[ -y + 3 = 0 \implies y = 3 \]

Now, substitute \( y = 3 \) back into Equation 1 to find \( x \):
\[ x - 3 = 1 \implies x = 4 \]

Thus, the intersection point \( P \) is \( (4, 3) \).

### Step 2: Find the slope of the line parallel to \( 3x + 4y = 12 \)
To find the slope of the line \( 3x + 4y = 12 \), we can rearrange it into slope-intercept form \( y = mx + c \):
\[ 4y = -3x + 12 \implies y = -\frac{3}{4}x + 3 \]
From this, we can see that the slope \( m \) of the line is \( -\frac{3}{4} \).

### Step 3: Use the point-slope form to find the equation of the desired line
We have the point \( P(4, 3) \) and the slope \( m = -\frac{3}{4} \). We can use the point-slope form of the equation of a line:
\[ y - y_0 = m(x - x_0) \]
Substituting \( (x_0, y_0) = (4, 3) \) and \( m = -\frac{3}{4} \):
\[ y - 3 = -\frac{3}{4}(x - 4) \]

### Step 4: Simplify the equation
Distributing the slope on the right side:
\[ y - 3 = -\frac{3}{4}x + 3 \]
Adding 3 to both sides:
\[ y = -\frac{3}{4}x + 6 \]

### Final Equation
Thus, the equation of the line through the intersection of the two lines and parallel to \( 3x + 4y = 12 \) is:
\[ y = -\frac{3}{4}x + 6 \]