The value(s) of *y* for which the following pair of equations

*x*

^{2}+

*y*

^{2}– 16 = 0 and

*x*

^{2}– 3

*y*+ 12 = 0

may have a real, common solution are:

- 4 only
- -7, 4
- 0, 4
- no
*y* - all
*y*.

We can graph the two equations — the circle and the parabola — to see what’s going on.

With algebra:

*x*

^{2}+

*y*

^{2}– 16 =

*x*

^{2}– 3

*y*+ 12

→

*y*

^{2}– 16 = -3

*y*+ 12

→

*y*

^{2}+ 3

*y*– 28 = 0

→ (

*y*+ 7)(

*y*– 4) = 0

→

*y*= 4 or -7.

Will 4 and -7 work in the two equations?

**4:**

*x*

^{2}+ 16 – 16 = 0

*x*= 0

OK.

*x*^{2} – 3 · 4 + 12 = 0

*x*^{2} – 12 + 12 = 0

*x* = 0

OK.

**-7:**

*x*

^{2}+ 49 – 16 ≟ 0

*x*

^{2}= -33

Nope.

*x*^{2} + 21 + 12 = 0

*x*^{2} = -33

Nope.

So we can use *y* = 4, and only 4. The answer is (a).