Solve the system of equations algebraically x^2+2y^2=10 3x^2-y^2=9
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We have to solve the system of equations:
x^2+2y^2=10...(1)
3x^2-y^2=9 ...(2)
(1) - 2*(2)
=> x^2 + 2y^2 - 3x^2 - 2y^2 = 10 - 18
=> -2x^2 = -8
=> x^2 = 4
Substitute in (2)
3*4 - y^2 = 9
=> y^2 = 3
x^2 = 4
=> x = -2 and 2
y^2 = 3
=> y = sqrt 3 and -sqrt 3
The required solutions are (2, sqrt 3),(2 , -sqrt 3), (-2 , sqrt 3), (-2, -sqrt 3)
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We'll multiply the second equation by 2:
6x^2 - 2y^2 = 18 (3)
We'll add the 3rd equation to the first one:
x^2 + 2y^2 + 6x^2 - 2y^2 = 10 + 18
We'll combine like terms:
7x^2 = 28
x^2 = 4
x1 = 2 and x2 = -2
Now, we'll substitute x = 2 in the 1st equation:
2^2 + 2y^2 = 10
4 + 2y^2 = 10
2y^2 = 6
y^2 = 3
y1 = sqrt3 and y2 = -sqrt3
We'll substitute x = -2 in the 1st equation:
4 + 2y^2 = 10
2y^2 = 6
y^2 = 3
y3 = sqrt3 and y4 = -sqrt3
The solutions of the equation are the ordered pairs: (2 ; sqrt3) ; (2 ; -sqrt3) ; (-2 ; sqrt3) ; (-2 ; -sqrt3).
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