🌆 Grade 9 · Lesson 12 of 12
Comet Interceptor: Solving Linear-Quadratic Systems
A straight path and a curved path can cross zero, one, or two times — substitution finds exactly where.
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The big idea: When a line and a parabola are graphed together, they can miss each other completely, touch at one point, or cross at two points. Substituting the linear expression for y into the quadratic equation collapses the system into a single quadratic equation in x — solve that, and you have found every point where the two paths meet.
Where a straight path meets a curved one
Picture a comet racing along a perfectly straight line while a space station follows a curved, parabolic orbit. Do their paths ever cross? If so, where — and how many times? Answering this is exactly what it means to solve a linear-quadratic system: find every point (x, y) that satisfies both equations at once.
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Substitute to collapse two equations into one
Since both equations describe y, set the linear expression for y equal to the quadratic expression for y. This replaces the system with a single quadratic equation in x, which you already know how to solve.
\[ y = mx + b \qquad y = ax^2 + bx + c \]
A generic linear-quadratic system: one straight-line equation and one parabola equation, both describing y in terms of x.
🎮 Comet Interceptor LIVE
Intersect a line with a parabola to find where the comet meets its trajectory.
From two equations to one
Set the two expressions for y equal, move everything to one side, and you have a standard quadratic equation in x. Factor it (or use the quadratic formula) to find the x-value(s) of every intersection point, then substitute each x back into either original equation to find the matching y-value.
📝 Worked example: Find every point where the line y = x + 1 meets the parabola y = x² − 1.
- Set the expressions for y equal: x + 1 = x² − 1.
- Move everything to one side: 0 = x² − x − 2, which factors as 0 = (x − 2)(x + 1).
- So x = 2 or x = −1. Substitute each back into y = x + 1: at x = 2, y = 3; at x = −1, y = 0.
✓ The line and parabola meet at <strong>(2, 3)</strong> and <strong>(−1, 0)</strong>.
📝 Worked example: Find every point where the line y = x meets the parabola y = x² − 2x.
- Set the expressions for y equal: x = x² − 2x.
- Move everything to one side: 0 = x² − 3x, which factors as 0 = x(x − 3).
- So x = 0 or x = 3. Substitute each back into y = x: at x = 0, y = 0; at x = 3, y = 3.
✓ The line and parabola meet at <strong>(0, 0)</strong> and <strong>(3, 3)</strong>.
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The discriminant predicts how many crossings
After substitution you get a single quadratic in x. Its discriminant (b² − 4ac) tells you how many real solutions exist before you even finish solving: positive means two intersection points, zero means exactly one (the line is tangent to the parabola), and negative means none — the line misses the parabola entirely.
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Always verify in both original equations
After solving, plug each (x, y) pair back into both original equations. A single sign slip while combining the equations can produce a solution that looks right but fails to actually satisfy one of the paths.
Check your understanding
1. Solve for the x-values where y = x − 1 meets y = x² − 3x + 2.
Set x − 1 = x² − 3x + 2, giving 0 = x² − 4x + 3 = (x − 1)(x − 3), so x = 1 or x = 3.
2. After substitution, a linear-quadratic system produces a quadratic with a negative discriminant. How many times do the line and parabola intersect?
A negative discriminant means the resulting quadratic has no real solutions, so the line and parabola never meet.
3. A linear-quadratic system's substitution produces a quadratic with a positive discriminant. How many intersection points are there?
A positive discriminant means two distinct real solutions for x, so the line and parabola cross at two points.
4. The point (2, 3) is proposed as a solution to the system y = x + 1 and y = x² − 1. Is it valid?
Check both: y = 2 + 1 = 3 ✓, and y = 2² − 1 = 3 ✓. The point satisfies both equations, so it is a valid intersection point.
5. What is the first algebra step to solve a system of one linear and one quadratic equation, both solved for y?
Since both equations equal y, setting them equal to each other collapses the system into one solvable quadratic equation in x.
✅ Key takeaways
- A linear-quadratic system asks where a straight line and a parabola meet, which can happen zero, one, or two times.
- Substituting one expression for y into the other collapses the system into a single quadratic equation in x.
- Solve that quadratic (by factoring or the quadratic formula) to get the x-values of every intersection, then find the matching y-values.
- The discriminant of the resulting quadratic predicts the number of intersection points before you finish solving.
- Always verify each solution point in both original equations to catch algebra errors.
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