Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4.

Solving x + 2y = 7 and x - y = 4 simultaneously,

⇒ x + 2y = 7

⇒ x = 7 - 2y …….(1)

Substituting above value of x in x - y = 4 we get,

⇒ 7 - 2y - y = 4

⇒ -3y = 4 - 7

⇒ -3y = -3

⇒ y = 1.

Substituting y = 1 in equation 1 we get,

⇒ x = 7 - 2(1) = 5.

Point of intersection = (5, 1).

Slope of line passing through (0, 0) and (5, 1) = $\dfrac{1 - 0}{5 - 0} = \dfrac{1}{5}.$

By point-slope form,

$y - y$1 = m(x - x1)

Substituting values we get,

⇒ y - 0 = $\dfrac{1}{5}(x - 0)$

⇒ 5(y - 0) = 1(x - 0)

⇒ 5y = x

⇒ x - 5y = 0.

Hence, equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4 is x - 5y = 0.