11.5 Solve Systems of Nonlinear Equations

Solving Systems of Nonlinear Equations

A nonlinear system of equations contains two or more equations where at least one equation isn't linear (it might be quadratic, circular, elliptical, etc.). Solving these systems means finding every point $(x, y)$ that satisfies all equations at once. Since this unit covers conics, you'll most often see systems combining lines, parabolas, circles, and ellipses.

The core methods are graphing, substitution, and elimination. Each works best in different situations, so knowing when to reach for which tool matters just as much as knowing how to use it.

Graphical Intersection for Nonlinear Systems

Graphing gives you a visual picture of how many solutions exist and roughly where they are. Unlike linear systems (which have at most one intersection), nonlinear systems can have zero, one, two, or even more intersection points.

How to use graphing:

1. Identify the type of each equation (line, parabola, circle, ellipse, hyperbola) so you know what shape to expect.
2. Graph both equations on the same coordinate plane. A graphing calculator or Desmos works well here.
3. Locate the points where the curves cross. Each intersection point is a solution to the system.
4. Estimate the coordinates of each intersection point from the graph.
5. Verify by plugging those coordinates back into both original equations.

Graphing is great for understanding the problem, but it often only gives approximate answers. For exact solutions, you'll need substitution or elimination.

Substitution Method for Nonlinear Systems

Substitution is usually your best bet for nonlinear systems, especially when one equation is already solved for a variable (or can be easily rearranged).

Step-by-step process:

1. Isolate a variable in whichever equation makes it easiest. Pick the equation and variable that avoid square roots or fractions when possible.
2. Substitute that expression into the other equation. You'll now have one equation with one variable.
3. Solve the resulting equation using whatever technique fits (factoring, quadratic formula, etc.).
4. Back-substitute each solution into the equation from Step 1 to find the other variable.
5. Check every solution pair in both original equations.

Example: Solve the system $y = x^2 + 1$ and $x + y = 5$.

• The first equation already has $y$ isolated, so substitute into the second: $x + (x^2 + 1) = 5$
• Simplify: $x^2 + x - 4 = 0$
• Use the quadratic formula: $x = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$
• Substitute each $x$-value back into $y = x^2 + 1$ to find the corresponding $y$-values.
• Check both pairs in the original equations.

Note that this system produces two solutions, which makes sense visually: a parabola and a line can intersect at zero, one, or two points.

Elimination Method for Nonlinear Systems

Elimination works well when both equations contain the same nonlinear term (like $x^2$ or $y^2$) and you can cancel it by adding or subtracting the equations.

Step-by-step process:

1. Align matching terms. Look for a squared term (or other nonlinear term) that appears in both equations.
2. Multiply one or both equations by constants so that the nonlinear terms have opposite coefficients.
3. Add or subtract the equations to eliminate that term. You should now have a simpler equation.
4. Solve the simpler equation for the remaining variable.
5. Substitute back into either original equation to find the other variable.
6. Verify all solution pairs in both original equations.

Example: Solve $x^2 + y = 7$ and $x^2 - y = 1$.

• Both equations contain $x^2$. Add them directly: $(x^2 + y) + (x^2 - y) = 7 + 1$ $2x^2 = 8$ $x^2 = 4$ $x = \pm 2$
• Substitute $x^2 = 4$ into the first equation: $4 + y = 7$, so $y = 3$.
• Solutions: $(2, 3)$ and $(-2, 3)$.
• Check: Both pairs satisfy $x^2 + y = 4 + 3 = 7$ ✓ and $x^2 - y = 4 - 3 = 1$ ✓.

Don't forget the $\pm$ when you take a square root. Missing the negative solution is one of the most common mistakes on these problems.

Choosing the Right Method

In practice, you'll often graph first to see what you're dealing with, then use substitution or elimination to get exact answers.

Real-World Applications of Nonlinear Systems

Many applied problems naturally produce nonlinear systems. The setup process is the same as for word problems with linear systems, but the equations involve squared terms or other nonlinear relationships.

General approach:

1. Define variables with clear meaning and units (e.g., let $x$ = horizontal distance in meters).
2. Write two equations based on the relationships described in the problem. These might come from geometry (area, the Pythagorean theorem), physics (projectile motion), or other constraints.
3. Solve using substitution or elimination.
4. Interpret and filter results. Not every algebraic solution makes sense in context. Negative distances or times, for instance, are usually not valid.

Example: A rectangular garden has a perimeter of 30 feet and an area of 50 square feet. Find its dimensions.

• Let $l$ = length and $w$ = width.
• Perimeter equation: $2l + 2w = 30$, which simplifies to $l + w = 15$.
• Area equation: $lw = 50$.
• From the first equation: $l = 15 - w$. Substitute into the second: $(15 - w)w = 50$.
• Expand: $15w - w^2 = 50$, or $w^2 - 15w + 50 = 0$.
• Factor: $(w - 5)(w - 10) = 0$, so $w = 5$ or $w = 10$.
• The dimensions are 5 ft by 10 ft (both solutions give the same rectangle).