The graphs of the equations

\(4x + \frac{1}{3}y = \frac{8}{3}\) and \(\frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0\) intersect at a point P. The point P also lies on the graph of the equation:

Given:

\(4x + \frac{1}{3}y = \frac{8}{3}\) and 

\(\frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0\)

Calculation:

\(4x + \frac{1}{3}y = \frac{8}{3}\) ....(1)

\(\frac{1}{2}x + \frac{3}{4}y + \frac{5}{2} = 0\)

⇒ \(\frac{1}{2}x + \frac{3}{4}y = -\frac{5}{2} \) .....(2)

Now,

Multiplying the number by 3 in equation (1), we get

⇒ 12x + y = 8 ....(3)

Again, multiplying the number by 24 in equation (2), we get

⇒ 12x + 18y = -60 ...(4)

Now,

From equation (3) and (4), we get

⇒ (18y – y) = (-60 – 8)

⇒ 17y = -68

⇒ y = -4

Now, Putting the value of y in equation (3), we get

⇒ 12x – 4 = 8

⇒ 12x = 12

⇒ x = 1

So, P(1, -4)

Now,

According to given option

x + 2y – 5 = 0

⇒ (1 – 8 – 5) = 0

⇒ -12 is not equal to 0

So, It is not satisfying the point P

Again,

3x – y – 7 = 0

⇒ (3 + 4 – 7) = 0

⇒ (7 – 7) = 0

⇒ 0 = 0

It is satisfying the point P

Again,

x – 3y – 12 = 0

⇒ (1 + 12 – 12) = 0

⇒ (13 – 12) = 0

⇒ 1 is not equal to 0

So, It is not satisfying the point P

Again,

4x – y + 7 = 0

⇒ (4 + 4 + 7) = 0

⇒ 15 is not equal to 0

So, It is not satisfying the point P

∴  The point P also lies on the graph of the equation is 3x – y – 7 = 0