The value(s) of y for which the following pair of equations
x2 + y2 – 16 = 0 and x2 – 3y + 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:
x2 + y2 – 16 = x2 – 3y + 12
→ y2 – 16 = -3y + 12
→ y2 + 3y – 28 = 0
→ (y + 7)(y – 4) = 0
→ y = 4 or -7.
→ y2 – 16 = -3y + 12
→ y2 + 3y – 28 = 0
→ (y + 7)(y – 4) = 0
→ y = 4 or -7.
Will 4 and -7 work in the two equations?
4:
x2 + 16 – 16 = 0
x = 0
OK.
x2 + 16 – 16 = 0
x = 0
OK.
x2 – 3 · 4 + 12 = 0
x2 – 12 + 12 = 0
x = 0
OK.
-7:
x2 + 49 – 16 ≟ 0
x2 = -33
Nope.
x2 + 49 – 16 ≟ 0
x2 = -33
Nope.
x2 + 21 + 12 = 0
x2 = -33
Nope.
So we can use y = 4, and only 4. The answer is (a).