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Solve each quadratic system. 2. 9x^2+4y^2=36 x^2-y^2=4

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laneymills | Student, Grade 11 | (Level 2) eNoter

Posted May 9, 2013 at 5:49 AM via web

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Solve each quadratic system.

2. 9x^2+4y^2=36

x^2-y^2=4

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted May 9, 2013 at 6:11 AM (Answer #1)

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We have

`9x^2+4y^2=36`   (i)

`x^2-y^2=4`        (ii)

We solve this system of quadratic equations by method of substitution.

so from (ii)

`x^2=4+y^2`       (iii) ,substitute  this value of `x^2`  in (i),thus we have

`9(4+y^2)+4y^2=36`

`36+9y^2+4y^2=36`

`36+13y^2=36`

`13y^2=0`

`y^2=0`

`y=0`    (iv)

substitute y=0 in (iii) , we have

`x^2=4+0`

`x^2=4`

`x=+-2`

Thus solution of system of equations are

`(+-2,0)`

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oldnick | (Level 1) Valedictorian

Posted May 9, 2013 at 7:02 AM (Answer #2)

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`9x^2+4y^2=36`   (1)

`x^2-y^2=4`          (2)

subtract at (1)  (2) multiplied by 9:

`9x^2+4y^2-9(x^2-y^2)=36-4xx9`

`13y^2=0`   `rArr y=0`

On the other side:

Adding at (1) (2) multiplied by 4:

`9x^2+4y^2+4(x^2-y^2)=36+4xx4 `

`13x^2=52`   `rArr` `y^2=4`  `rArr x=+-2`

Solutions are   `(+-2;0)`  

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