Question

Solving simultaneous equations with one quadratic, x^2+y^2=16 and y=x+3

Answer 1

If
`y=x+3`, then in the first equation you have


`x^2+(x+3)^2=x^2+(x^2+6x+9)=2x^2+6x+9=16`

Rewriting a bit, you get


`x^2+3x+\frac92=8`


`x^2+3x+\frac94+\frac94=8`


`\left(x+\frac32\right)^2=\frac{23}4`


`x=-\frac32\pm\frac{√(23)}2`

Now, since
`y=x+3`, you also get


`y=-\frac32\pm\frac{√(23)}2+3=\frac32\pm\frac{√(23)}2`

So, there are two solutions here:


`(x,y)=\left(-\frac32+\frac{√(23)}2,\frac32+\frac{√(23)}2\right)`

`(x,y)=\left(-\frac32-\frac{√(23)}2,\frac32-\frac{√(23)}2\right)`