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.