Nonlinear Systems of Equations

A system of equations is a set of two or more equations that you solve together. A nonlinear system is one where at least one of the equations is not a straight line—such as a parabola ($y = x^2$), a circle ($x^2 + y^2 = r^2$), or another curve.

Geometrically, solving a nonlinear system means finding the exact coordinate points where these graphs intersect.

The Substitution Method

The most reliable way to solve nonlinear systems is the substitution method. Follow these steps:

1. Isolate one variable (like $y$ or $x$) in the simpler equation (usually the linear one).
2. Substitute that expression into the other, nonlinear equation.
3. Solve the new equation, which will now only have one variable. (This often requires solving a quadratic equation).
4. Plug back your solutions into the isolated equation from Step 1 to find the corresponding values for the other variable.

Example 1: A Line and a Parabola

Problem: Solve $y = x^2 - 3x + 2$ and $y = x - 1$ simultaneously.

Step 1: Both equations are already solved for $y$. Step 2: Substitute the linear expression for $y$ into the quadratic equation: $x - 1 = x^2 - 3x + 2$

Step 3: Move all terms to one side to set the equation to zero: $0 = x^2 - 4x + 3$

Factor the quadratic equation: $0 = (x - 1)(x - 3)$

This gives us two $x$-values: $x = 1$ and $x = 3$.

Step 4: Plug these $x$-values back into the linear equation ($y = x - 1$) to find the $y$-values:

• If $x = 1$, then $y = 1 - 1 = 0$. So, the first intersection point is $(1, 0)$.
• If $x = 3$, then $y = 3 - 1 = 2$. So, the second intersection point is $(3, 2)$.

Solution: The intersection points are $(1, 0)$ and $(3, 2)$.

Example 2: Two Nonlinear Equations

Problem: Find the intersection points of $y = x^2$ and $x^2 + y^2 = 20$.

Step 1 & 2: The first equation already has $y$ isolated in terms of $x^2$. We can substitute $y$ directly in place of $x^2$ in the second equation: $y + y^2 = 20$

Step 3: Rearrange into a standard quadratic form: $y^2 + y - 20 = 0$

Factor the equation: $(y + 5)(y - 4) = 0$

This gives $y = -5$ and $y = 4$.

Step 4: Plug these $y$-values back into $y = x^2$ to find $x$:

• If $y = -5$: $-5 = x^2$. Since the square of a real number cannot be negative, there are no real solutions for this branch.
• If $y = 4$: $4 = x^2$. Taking the square root gives $x = 2$ and $x = -2$.

Solution: The intersection points are $(2, 4)$ and $(-2, 4)$.