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)
