Two curves can meet nought, once, twice, three times or four. The algebra must agree with the picture.
Solve the system
x² + y² = 25 and y = x + 1
Substitute: x² + (x + 1)² = 25 gives 2x² + 2x − 24 = 0, so x² + x − 12 = 0 and (x + 4)(x − 3) = 0. Then x = 3 gives y = 4, and x = −4 gives y = −3 — both points. A line cutting a circle meets it twice, so stopping at one answer leaves half the solution behind.
y = x + 1 into x² + y² = 13
the line is already solved for y
x² + (x + 1)² = 13
one variable left
2x² + 2x − 12 = 0
a quadratic, as expected
x = 2 or x = −3
two x values, two points
(2, 3) and (−3, −2)
y = x + 1 for each
Every x must be turned back into a point. An answer of "x = 2 or x = −3" is only half done; the question asked where the curves meet.
x² + y² = 34, x² − y² = 16
a circle and a hyperbola
add: 2x² = 50, x² = 25
the y² terms cancel
x = ±5, then y² = 9
back-substitute
(5, 3), (5, −3), (−5, 3), (−5, −3)
four points
Pairing them up as (5, 3) and (−5, −3) and stopping loses half the answer. Here all four combinations satisfy both equations, because both contain only squares and are therefore symmetric in both axes.
A quick sketch is worth the thirty seconds. If the algebra returns three points for a line and a circle, the sketch will tell you immediately that something has gone wrong.
x² = 4x − 4
substitute
x² − 4x + 4 = 0
discriminant 16 − 16 = 0
(x − 2)² = 0, x = 2
one value, twice over
a single point (2, 4)
the line is tangent
A zero discriminant is the algebra's way of saying "tangent". A negative one says the curves never meet, and "no solution" is then the correct and complete answer.
The same two moves, substitute or eliminate, with three different endings.
No solution: x² + y² = 4 and y = x + 4
x² + (x + 4)² = 4
substitute the line
2x² + 8x + 12 = 0, so x² + 4x + 6 = 0
divide by 2
discriminant 16 − 24 = −8
negative
no solution
the line misses the circle
A negative discriminant is a complete answer, not a dead end.
Subtract to eliminate: x² + 4y² = 20 and x² + y² = 8
3y² = 12, so y² = 4
first minus second; x² cancels
x² = 8 − 4 = 4
back-substitute
(2, 2), (2, −2), (−2, 2), (−2, −2)
every sign combination
When the x² coefficients match, subtract rather than add.
Three points: y = x² − 3 and x² + y² = 9
x² = y + 3, so (y + 3) + y² = 9
substitute for x², not for y
y² + y − 6 = 0, so y = 2 or y = −3
factor (y + 3)(y − 2)
y = 2: x² = 5, x = ±√5; y = −3: x² = 0, x = 0
back to x² = y + 3
(√5, 2), (−√5, 2), (0, −3)
the parabola's vertex touches the circle
Two conics can share 0, 1, 2, 3 or 4 points. Here y = −3 gives only one x, so the count is three.
y = 7 − x into xy = 10
substitute
x(7 − x) = 10, so x² − 7x + 10 = 0
tidy into standard form
x = 2 or x = 5
two candidates
(2, 5) and (5, 2), both check ✓
2 × 5 = 10 and 2 + 5 = 7
Squaring or multiplying through can invent roots that satisfy the new equation but not the original, exactly as with radical and rational equations. Substituting each answer back into both starting equations is the only way to be sure.
Tests give a line and a circle or two conics and ask for the intersection points, or only for how many there are. The fastest route: substitute when one equation is a line, add or subtract when both have x² and y², and turn every value back into a point. A 10-second check: put each point into both original equations.
Count the solutions
x² + y² = 20 and x² − y² = 12
Adding gives 2x² = 32, so x² = 16 and x = ±4. Then y² = 4, so y = ±2. Every combination works, giving 4 points: (4, 2), (4, −2), (−4, 2) and (−4, −2). B pairs the signs up and reports only two.
Hint: both equations contain only squares, so the picture is symmetric about both axes. Whatever happens in one quadrant happens in all four.
