9.3Nonlinear Systems and Inequality Regions
Solve nonlinear systems (line-parabola, circle-line, parabola-parabola) by substitution and elimination. Graph systems of inequalities and shade feasible regions.
Nonlinear systems arise when modeling intersections of curves — where a line meets a parabola, or two circles overlap. Systems of inequalities define feasible regions used in linear programming and optimization.
Essential Question
How do you find the intersection points of a line and a curve (or two curves), and how do you graph the solution region of a system of nonlinear inequalities?
Lesson Overview
A nonlinear system contains at least one equation that is not linear — such as a parabola, circle, or hyperbola. Unlike linear systems, nonlinear systems can have 0, 1, 2, or more intersection points.
Methods for Solving
- •Substitution: solve one equation for a variable, then substitute into the other equation.
- •Elimination: add or subtract equations to eliminate a variable — works well when both equations contain x² or y² terms.
Systems of Inequalities
- 1.Graph each inequality boundary (solid curve for ≤/≥, dashed for </>)
- 2.Shade the solution region for each inequality separately.
- 3.The solution to the system is the intersection of all shaded regions.
- 4.Test a point (usually the origin) to determine which side to shade.
Common Nonlinear Systems
Worked Examples
Solve the system: y = x² and y = x + 2.
Set the expressions equal: x² = x + 2
Rearrange: x² − x − 2 = 0
Factor: (x − 2)(x + 1) = 0
Solve: x = 2 or x = −1
Find y: when x = 2, y = 4; when x = −1, y = 1
Solve the system: x² + y² = 25 and y = x + 1.
Substitute y = x + 1 into the circle: x² + (x + 1)² = 25
Expand: x² + x² + 2x + 1 = 25
Simplify: 2x² + 2x − 24 = 0 → x² + x − 12 = 0
Factor: (x + 4)(x − 3) = 0
x = −4 → y = −3; x = 3 → y = 4
Solve the system: x² + y² = 10 and x² − y² = 2.
Add the two equations: 2x² = 12 → x² = 6 → x = ±√6
Substitute back: y² = 10 − 6 = 4 → y = ±2
All combinations of signs are valid solutions.
Graph the system of inequalities: y ≥ x² and y ≤ 4.
Graph y = x² (parabola, solid boundary — ≥ is non-strict).
Graph y = 4 (horizontal line, solid boundary — ≤ is non-strict).
Shade above the parabola (y ≥ x²) and below y = 4 (y ≤ 4).
The solution region is the area between the parabola and the line.
Check: (0, 2) → 2 ≥ 0 ✓ and 2 ≤ 4 ✓ — origin is in the region.
Solve the system: y = x² − 3 and y = 2x.
Set equal: x² − 3 = 2x
Rearrange: x² − 2x − 3 = 0
Factor: (x − 3)(x + 1) = 0
x = 3 → y = 6; x = −1 → y = −2
Guided Practice
Solve: y = x² − 4 and y = 2x − 1.
Hint: Set the expressions equal and solve the resulting quadratic.
Solve: x² + y² = 13 and x + y = 5.
Hint: Solve the linear equation for y, then substitute into the circle equation.
Solve: y = x² and y = −x² + 8.
Hint: Set the two expressions equal — both equal y.
Determine the number of solutions: y = x² + 1 and y = x − 1.
Hint: Set equal and check the discriminant of the resulting quadratic.
Graph the system: y ≤ x² and y ≥ −x + 2.
Hint: Graph each boundary, test a point in each region, shade the intersection.
Key Vocabulary
Nonlinear system
A system where at least one equation is not linear — it may contain squared terms, products of variables, or other nonlinear expressions.
Substitution method
Solving one equation for a variable and substituting that expression into the other equation to reduce the system to one equation in one variable.
Elimination method
Adding or subtracting equations to eliminate a variable. Especially useful when both equations contain squared terms (x² or y²).
System of inequalities
Two or more inequalities considered simultaneously. The solution is the region satisfying all of them at once.
Solution region
The set of all points (x, y) that satisfy every inequality in the system. It is the intersection of all individual shaded regions.
Boundary curve
The curve (line, parabola, circle, etc.) that forms the edge of a solution region. Drawn solid for ≤ or ≥; drawn dashed for < or >.
Test point
A point substituted into an inequality to determine which side of the boundary to shade. The origin (0, 0) is the most convenient choice when it does not lie on the boundary.
Quick Check
Interactive Practice — 5 Questions
How many solutions can a line-parabola system have?
Solve: y = x² and y = 4.
For the system y ≤ x + 3 and y ≥ x² − 1, the solution region is:
Solve: x² + y² = 25 and x = 3.
The boundary line y = 2x + 1 is dashed when the inequality is:
Common Mistakes
Only finding one intersection point when two exist.
Always solve the full quadratic. Factor or use the quadratic formula to find ALL roots — each root gives an intersection point.
Forgetting to find the y-coordinate after finding x.
A solution is an ordered pair (x, y). After finding x, substitute back into either original equation to find y.
Shading the wrong region for an inequality.
Always test a point (like the origin) in the original inequality. If it satisfies the inequality, shade that side.
Using a solid boundary for a strict inequality (< or >).
Strict inequalities (< or >) use a dashed boundary. Non-strict (≤ or ≥) use a solid boundary.
Math Tips
For line-parabola systems, substitution is almost always the best method — substitute the linear equation into the quadratic.
For two-circle or two-parabola systems, elimination often works better — subtract the equations to cancel the squared terms.
The discriminant b² − 4ac tells you how many solutions a line-parabola system has: positive → 2 solutions, zero → 1 solution, negative → 0 solutions.
When graphing a system of inequalities, shade each region separately first, then identify the overlap.
The origin (0, 0) is the easiest test point — unless the boundary passes through the origin, in which case choose (1, 0) or (0, 1).