Question: Find the equation of the line passing through the point of intersection of the lines 3x +y = 7 and 3y = 4x -5 and parallel to the line 2x -y = 3.

Solution:

Line 1
Equation of the line is 2x - y = 3
Slope of line (m_1) = - \frac{coefficient \; of \; x}{ coefficient\; of y}
= - \frac{2}{-1}
= 2

Given,
Second line passes through 3x +y = 7 and 2y = 4x -5.
Solving the equations simultaneously,

3x + y = 7
or, y = 7 - 3x - (i)

Also,
3y = 4x -5
[Put value of y from equation (i) and solve]
or, 3(7 -3x) = 4x -5
or, 21 - 9x = 4x -5
or, 21+5 = 4x +9x
or, 26 = 13x
or, x = \frac{26}{13}
\therefore x = 2

Put value of x in equation (i), we get,
or, y = 7-3×2
or, y = 7-6
\therefore y = 1

(x,y) = (2,1)

So, coordinates of the point on the line 2 is (2,1).

According to the question,
Two lines are parallel to each other. Let m_2 represent the slope of the second line.
We have,
m_1 = m_2
or, 2 = m_2
\therefore m_2 = 2

Equation of line 2 when m_2 = 2 and point (2,1) = (x_1,y_1) is given by the formula,
(y-y_1) = m_2 (x -x_1)
or, (y -1) = 2(x -2)
or, y -1 = 2x -4
or, 2x -y -4+1 = 0
or, 2x - y -3 = 0 is the required equation.

Therefore, the required equation of the line is 2x -y -3 = 0.