Find the equation of a straight line passing through the intersection of 2x + 5y - 4 = 0 with x-axis and parallel to the line 3x - 7y + 8 = 0.

Let the point of intersection of the line 2x + 5y - 4 = 0 and the x-axis be (x₁, 0).

Substituting the value of points in equation,

⇒ 2x₁ + 5 × 0 - 4 = 0
⇒ 2x₁ = 4
⇒ x₁ = 2.

Coordinates of the point of intersection will be (2, 0).

Given new line is parallel to 3x - 7y + 8 = 0.

Converting it in the form y = mx + c,

3x - 7y + 8 = 0

⇒ 7y = 3x + 8

⇒ y = $\dfrac{3}{7}x + \dfrac{8}{7}$

Comparing equation with y = mx + c, we get slope = $\dfrac{3}{7}$.

The equation of the line with slope $\dfrac{3}{7}$ and passing through (2, 0) can be given by point-slope form,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 0 = \dfrac{3}{7}(x - 2) \\[1em] \Rightarrow 7y = 3x - 6 \\[1em] \Rightarrow 3x - 7y - 6 = 0.

Hence, the equation of the new line is 3x - 7y - 6 = 0.