Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

x – 7y + 5 = 0 …(i)

3x + y – 7 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (i) by 3, we get

3x – 21y + 15 = 0 …(iii)

On subtracting eq. (iii) from (ii), we get

3x + y – 7 – 3x + 21y – 15 = 0

⇒ 22y - 22 = 0

⇒ 22y = 22

⇒ y = 1

Putting the value of y in eq. (i), we get

x – 7(1) + 5 = 0

⇒ x – 7 + 5 = 0

⇒ x – 2 = 0

⇒ x = 2

Hence, the point of intersection P(x₁, y₁) is (2, 1)

The equation of line parallel to x – axis is of the form

y = b where b is some constant

Given that this equation of the line passing through the point of intersection (2, 1)

Hence, point (2, 1) will satisfy the equation of a line.

Putting y = 1 in the equation y = b, we get

y = b

⇒ 1 = b

or b = 1

Now, the required equation of a line is y = 1