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

### 2 Answers | Add Yours

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)`

`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)`