Non-Linear Systems
A non-linear system of equations is a set of two or more equations in which at least one equation is not linear — meaning at least one variable is raised to a power other than one, appears inside a function, or is multiplied by another variable. The most common type you will encounter at this level is the linear-quadratic system, where one equation describes a straight line and the other describes a parabola. These systems arise naturally whenever a moving object's curved path intersects a boundary, or when you need to find where a line meets a circle or curve. Unlike purely linear systems, which have either exactly one solution, no solution, or infinitely many solutions (all lying on a line), a linear-quadratic system can have zero, one, or two solutions. Geometrically, two solutions appear when a line crosses a parabola at two distinct points, one solution appears when the line is tangent to the parabola (touching it at exactly one point), and no real solution appears when the line misses the parabola entirely. Understanding this geometric picture helps you anticipate how many answers to expect before you begin solving. The primary algebraic technique for solving these systems is substitution: you express one variable from the linear equation and substitute it into the non-linear equation, reducing the problem to a single-variable polynomial equation that you can solve using factoring, completing the square, or the quadratic formula. You can also use elimination in some cases, particularly when both equations share a squared term that can be cancelled. Checking your answers by substituting back into both original equations is essential, because the algebra can sometimes introduce extraneous solutions.
- Write both equations clearly
- Substitute the linear expression into the quadratic equation
- Rearrange into standard quadratic form
- Solve the quadratic equation
- Find the corresponding y-values
- Verify your solutions
Write both equations clearly
Identify which equation is linear (degree 1) and which is quadratic (degree 2). For example, y = 2x + 1 is linear and y = x² − 3 is quadratic. Make sure each equation is solved for the same variable if possible, as this makes substitution straightforward.
Substitute the linear expression into the quadratic equation
Replace the variable in the quadratic equation with the expression from the linear equation. For instance, substitute y = 2x + 1 into y = x² − 3 to get 2x + 1 = x² − 3. This produces a single quadratic equation in x alone.
Rearrange into standard quadratic form
Move all terms to one side so the equation equals zero: x² − 2x − 4 = 0. The standard form is ax² + bx + c = 0, where a ≠ 0. Identify a, b, and c carefully before proceeding.
Solve the quadratic equation
Use factoring if the discriminant yields perfect-square factors, or apply the quadratic formula x = (−b ± √(b² − 4ac)) / (2a). The discriminant Δ = b² − 4ac tells you the number of real solutions: Δ > 0 gives two distinct x-values, Δ = 0 gives one repeated x-value, and Δ < 0 means no real intersection exists.
Find the corresponding y-values
For each x-value found, substitute back into the simpler (linear) equation to find the matching y-value. This gives you the solution pairs (x, y). Never substitute into only one equation and forget the other — always verify both coordinates satisfy both original equations.
Verify your solutions
Substitute each solution pair into both original equations to confirm they satisfy them. Write your final answers as ordered pairs, for example (x₁, y₁) and (x₂, y₂), and describe them geometrically as the intersection points of the line and the parabola.
The Key Equation
- x — solution variable (none): The unknown value(s) of x that satisfy the quadratic equation ax² + bx + c = 0.
- a — leading coefficient (none): The coefficient of x², which must be non-zero. Determines the direction and width of the parabola.
- b — linear coefficient (none): The coefficient of x. Together with a and c, determines the position of the vertex and the roots.
- c — constant term (none): The term independent of x. Represents the y-intercept of the quadratic when written as y = ax² + bx + c.
- \Delta = b^2 - 4ac — discriminant (none): Determines the number of real solutions: positive means two real roots, zero means one repeated root, negative means no real roots.
Worked Example
Problem. Solve the system: y = x + 3 and y = x² − 2x + 1. Find all intersection points.
Answer. Two intersection points: x = (3 ± √17)/2 with corresponding y = (9 ± √17)/2.
Origins · 1 minRené Descartes and how this was discovered
Descartes invented the Cartesian coordinate system, published in his 1637 work La Géométrie. By assigning numerical coordinates to geometric points, he unified algebra and geometry, making it possible to represent equations like lines and parabolas as curves on a plane and to find their intersections algebraically — the very foundation of solving non-linear systems graphically and algebraically. Descartes reportedly stayed in bed until noon each day to do his best thinking. Legend holds that the idea for the coordinate system came from watching a fly crawl across a tiled ceiling and realizing its position could be described by two numbers.
Related Concepts
- Substitution Method for Linear Systems — Non-linear systems are solved primarily by substitution; mastering it for linear systems first builds the essential skill needed here.
- The Quadratic Formula — After substitution reduces a linear-quadratic system to a single quadratic equation, the quadratic formula is often the tool used to solve it.
- Discriminant and Nature of Roots — The discriminant tells you in advance how many intersection points a linear-quadratic system will have — zero, one, or two.
- Graphing Quadratic Functions — Understanding parabolas geometrically deepens your intuition for why a line and a parabola can intersect in 0, 1, or 2 places.
- Pair of Linear Equations — Non-linear systems extend the methods developed for pairs of linear equations; comparing the two helps highlight what changes and what stays the same.