Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y – 3) + k(2x – y + 1) = 0 …..(i)
x + y – 3 + 2kx – ky + k = 0
x + 2kx + y – ky – 3 + k = 0
(1 + 2k)x + (1 – k)y – 3 + k = 0
But, this line is parallel to X-axis Its slope = 0
⇒ \(\frac {-(1+2k)}{1-k} =0\)
⇒ 1 + 2k = 0
⇒ k = -1/2
Substituting the value of k in (i), we get
(x + y – 3) + -1/2 (2x – y + 1) = 0
⇒ 2(x + y – 3) – (2x – y + 1 ) = 0
⇒ 2x + 2y – 6 – 2x + y – 1 = 0
⇒ 3y – 7 = 0, which is the equation of the required line.
