Question

Question #df17c

Answer 1

By raising `6` to both sides of the equations,

`{(3x^2+y^2=28),(x+y=6 => y=6-x):}`

by substituting the second equation into the first equation,

`3x^2+(6-x)^2=28 => 4x^2-12x+8=0`

by dividing by `4`,

`=> x^2-3x+2=0`

by factoring out,

`=>(x-1)(x-2)=0 => x=1,2 => y=5,4`

Hence, the solutions are

`(x,y)=(1,5)` and `(2,4)`

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I hope that this was helpful.