Find the real solution of the system 3x^2-2y^2=-5 xy=12

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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This is a homogenous system and the first step is to eliminate the numbers alone.

For this purpose, we'll multiply the 1st equation by 12 and the 2nd equation by 5:

36x^2 - 24y^2 = -60 (3)

5xy = 60 (4)

We'll add (3) and (4):

36x^2 + 5xy - 24y^2 = 0

We'll divide by x^2:

36 + 5y/x - 24y^2/x^2 = 0

We'll substitute y/x = t:

-24t^2 + 5t + 36 = 0

24t^2 - 5t - 36 = 0

We'll apply quadratic formula:

t1 = [5+sqrt(25+3456)]/48

t1 = (5+59)/48

t1 = 4/3

t2 = (5-59)/48

t2 = -9/8

We'll put y/x = 4/3

y = 4x/3

We'll substitute y in the 2nd equation:

x*(4x/3) = 12

4x^2 = 4*9

x^2 = 9

x1 = 3 and x2 = -3

y1 = 4 and y2 = -4

y/x = -9/8

y = -9x/8

-9x^2 = 8*12

-3x^2 = 32

x^2 = -32/3

This equation has no real solutions.

The system will have the following real solutions {(3 ; 4) ; (-3 ; -4)}.

 

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