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Zepler [3.9K]
3 years ago
5

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

for me.... ​

Mathematics
1 answer:
kotegsom [21]3 years ago
8 0

That's not a linear system, but you have an awesome school system for giving you this problem.

\dfrac 5 y - \dfrac 2 x = \dfrac{13}6

Multiply by 6xy to clear the fractions.

30x - 12y = 13 xy

That's a second degree equation, also known as a conic.  That one happens to be a hyperbola.  

30x = y(13 x +12)

y = \dfrac{30x}{13 x + 12}

Let's clear the fractions from the second equation, multiplying out common denominator xy:

\dfrac {36} x - \dfrac {24} y = 1

36y - 24 x = xy

We are being asked to find the meet of two hyperbolas, so we expect two answers, a quadratic equation.

Substituting,

36 \left( \dfrac{30x}{13 x + 12} \right) - 24 x = x \left( \dfrac{30x}{13 x + 12} \right)

36(30x) -24 x(13x + 12) = 30x^2

1080 x - 312 x^2- 288 x = 30x^2

x(792 - 342 x)= 0

x = 0 \textrm{ or } x=792/342 = \dfrac{44}{19}

We have to rule out x=0 because it's in the denominator.

y = \dfrac{30x}{13 x + 12} = \dfrac{30(44/19)}{13(44/19)+12}

y = \dfrac{33}{20}

Answer: (44/19, 33/20)

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