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Solve the system of equations algebraically x^2+2y^2=10 3x^2-y^2=9

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noralbbig | Student, Grade 10 | eNoter

Posted February 22, 2011 at 2:53 AM via web

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Solve the system of equations algebraically

x^2+2y^2=10

3x^2-y^2=9

Tagged with algebra1, math

2 Answers | Add Yours

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giorgiana1976 | College Teacher | Valedictorian

Posted February 22, 2011 at 2:58 AM (Answer #1)

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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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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 22, 2011 at 3:04 AM (Answer #2)

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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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