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

∴ 4x + 3y = 5

or 4x + 3y – 5 = 0 …(i) and x = 2y – 7

or x – 2y + 7 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii) Multiply the eq. (ii) by 4, we get

4x – 8y + 28 = 0 …(iii)

On subtracting eq. (iii) from (i), we get 4x – 8y + 28 – 4x – 3y + 5 = 0

⇒ -11y + 33 = 0

⇒ -11y = -33

Putting the value of y in eq. (i), we get 4x + 3(3) – 5 = 0

⇒ 4x + 9 – 5 = 0

⇒ 4x + 4 = 0

⇒ 4x = -4

⇒ x = -1

Hence, the point of intersection P(x1, y1) is (-1, 3)