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



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pramodpandey's profile pic

Posted on (Answer #1)

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






`y=0`    (iv)

substitute y=0 in (iii) , we have




Thus solution of system of equations are


oldnick's profile pic

Posted on (Answer #2)

`9x^2+4y^2=36`   (1)

`x^2-y^2=4`          (2)

subtract at (1)  (2) multiplied by 9:


`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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