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Find when a parameterized system has exactly two solutions

Count distinct intersections, not just algebraic branches

A product equals zero when at least one factor is zero, but the domain must be respected. If two branches give the same ordered pair, they count as one solution.

Problem

For which real values of $a$ does the system $⎩ ⎨ ⎧ \frac{( y ^{2} - x y + 3 x - y - 6 ) x + 2}{6 - x} = 0, x + y = a$ have exactly two distinct real solutions $(x, y)$ ?

Answer

a \in (- 6, 1] \cup {8} \cup [9, 10)

The endpoint $a=-6$ is excluded: two apparent branches meet at the same point $(-2,-4)$.

Step-by-step solution

1. Find the domain and factor

The square roots require $- 2 \leq x < 6$ . The polynomial factor is

y^{2} - x y + 3 x - y - 6 = (y - 3) (y + 2 - x) .

Within the domain, the first equation is true on three branches: $x = - 2$ , $y = 3$ , or $y = x - 2$ .

2. Intersect each branch with $x + y = a$

The branch $x = - 2$ gives $(- 2, a + 2)$ for every $a$ . The branch $y = 3$ gives $(a - 3, 3)$ only when $1 \leq a < 9$ . The branch $y = x - 2$ gives $((a + 2) /2, (a - 2) /2)$ only when $- 6 \leq a < 10$ .

3. Count distinct pairs

At $a = - 6$ , the first and third branches both give $(- 2, - 4)$ , so there is only one solution. For $- 6 < a < 1$ there are two. At $a = 1$ , the first and second branches meet, while the third gives another solution. For $1 < a < 9$ , there are normally three, except at $a = 8$ , when the second and third meet at $(5, 3)$ . At $9 \leq a < 10$ , the $y = 3$ branch is outside the domain, leaving two. For $a \geq 10$ , only the first branch remains.

Therefore the required set is $(- 6, 1] \cup {8} \cup [9, 10)$ .

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