Can anyone plzzzz help me with this question of linear equation in two varaiable plzzzzzz... Its very important... Plz solve it for me.... ​

That's not a linear system, but you have an awesome school system for giving you this problem.[tex] \dfrac 5 y - \dfrac 2 x = \dfrac{13}6 [/tex]Multiply by 6xy to clear the fractions.[tex] 30x - 12y = 13 xy [/tex]That's a second degree equation, also known as a conic.  That one happens to be a hyperbola.  [tex] 30x = y(13 x +12)[/tex][tex] y = \dfrac{30x}{13 x + 12}[/tex]Let's clear the fractions from the second equation, multiplying out common denominator xy:[tex] \dfrac {36} x - \dfrac {24} y = 1 [/tex][tex] 36y - 24 x = xy[/tex]We are being asked to find the meet of two hyperbolas, so we expect two answers, a quadratic equation.Substituting,[tex]36 \left( \dfrac{30x}{13 x + 12} \right) - 24 x = x \left( \dfrac{30x}{13 x + 12} \right)[/tex][tex]36(30x) -24 x(13x + 12) = 30x^2[/tex][tex]1080 x - 312 x^2- 288 x = 30x^2[/tex][tex]x(792 - 342 x)= 0[/tex][tex] x = 0 \textrm{ or } x=792/342 = \dfrac{44}{19}[/tex]We have to rule out x=0 because it's in the denominator.[tex]y = \dfrac{30x}{13 x + 12} = \dfrac{30(44/19)}{13(44/19)+12}[/tex][tex]y = \dfrac{33}{20}[/tex]Answer: (44/19, 33/20)