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

$x^2+3x+\dfrac92=8$

$x^2+3x+\dfrac94+\dfrac94=8$

$\left(x+\dfrac32\right)^2=\dfrac{23}4$

$x=-\dfrac32\pm\dfrac{\sqrt{23}}2$

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

$y=-\dfrac32\pm\dfrac{\sqrt{23}}2+3=\dfrac32\pm\dfrac{\sqrt{23}}2$

So, there are two solutions here:

$(x,y)=\left(-\dfrac32+\dfrac{\sqrt{23}}2,\dfrac32+\dfrac{\sqrt{23}}2\right)$
$(x,y)=\left(-\dfrac32-\dfrac{\sqrt{23}}2,\dfrac32-\dfrac{\sqrt{23}}2\right)$