Question

plz help me........ bagwan tera bhala karega...pls​

Answer 1

it is too complicated to solve straight with division values to we will let p and q as the division value. it will make the problem easy to solve. check this photo. Hope it’ll be helpful ☺️

Answer 2

Consider the first equation,

$\frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \\ \\ \frac{(2 \times 3y) + 2x}{3xy} = \frac{1}{6} = \\ \\ \frac{6y + 2x}{3xy} = \frac{1}{6} \\ \\ 6y + 2x = \frac{3xy}{6} \\ \\ 6y + 2x = \frac{xy}{2} \\ \\ 2(6y + 2x) = xy \\ \\ 12y + 4x = xy$

Let the above equation be equation 1

Consider the second equation,

$\frac{3}{x} + \frac{2}{y} = 0 \\ \\ \frac{3y + 2x}{xy} = 0 \\ \\ 3y + 2x = 0 \\ \\ 3y = - 2x \\ \\ y = - \frac{2x}{3}$

Let the above equation be equation 2

Substitute y = - 2x/3 in equation 1

$12y + 4x = xy \\ \\ (12 \times ( \frac{ - 2x}{3} )) + 4x = x \times ( \frac{ - 2x}{3} ) \\ \\ (4 \times ( - 2x)) + 4x = \frac{ - 2 {x}^{2} }{3} \\ \\ - 8x + 4x = \frac{ - 2 {x}^{2} }{3} \\ \\ - 4x = \frac{ - 2 {x}^{2} }{3} \\ \\ - 4x \times 3 = - 2 {x}^{2} \\ \\ 12x = 2 {x}^{2} \\ \\ \frac{12x}{2} = {x}^{2} \\ \\ 6x = {x}^{2} \\ \\6 = \frac{ {x}^{2} }{x} \\ \\ x = 6$

Substitute x= 6 in equation 2.

$y = \frac{ - 2x}{3} \\ \\ y = \frac{ - 2 \times 6}{3} \\ \\ y = - 2 \times \frac{6}{3} \\ \\ y = - 2 \times 2 \\ \\ y = - 4$

Consider the third equation from question,

$y = ax - 4 \\ - 4 =( a \times 6) - 4 \\ \\ - 4 + 4 = 6a \\ \\ 6a = 0 \\ \\ a = 0$

The value of a is 0.