Question

How do you solve the system `2x^2+2y^2=15` and `x+2y=6`?

Answer 1

`x=0.72; y=2.64`
`x=1.68; y=2.16`

Explanation:

Given -

`2x^2+2y^2=15`-----------(1)
`x+2y=6` ---------------(2)

Solve equation (2) for `x`

`x=6-2y`

Substitute `x=6-2y` in equation (1)

`2(6-2y)^2+2y^2=15`
`2(36-24y+4y^2)+2y^2=15`
`72-48y+8y^2+2y^2=15`
`10y^2-48y+72-15=0`
`10y^2-48y+57=0`

`y` has two values

`y=(-b+-sqrt(b^2-(4xxaxxc)))/(2xxa`

`y=(-(-48)+-sqrt((-48)^2-(4xx10xx57)))/(2xx10`

`y=(48+-sqrt(2304-2280))/(2xx10`
`y=(48+-sqrt(24))/(2xx10)`

`y=(48+-4.9)/(20)`
`y=(48+4.9)/(20)=52.9/20=2.64`
`y=2.64`
`y=(48-4.9)/(20)=43.1/20=2.16`
`y=2.16`

`x` also has two values

Substitute `y=2.64` in equation (1)

`x+2(2.64)=6`
`x+5.28=6`
`x=6-5.28=0.72`
`x=0.72`

Substitute `y=2.16` in equation (1)

`x+2(2.16)=6`
`x+4.32=6`
`x=6-4.32=1.68`
`x=1.68`

`x,y` values are -
`x=0.72; y=2.64`
`x=1.68; y=2.16`