Lesson 3: Nonlinear Systems of Equations
Learning Objectives
- Understand what makes a system nonlinear
- Determine the possible number of solutions for nonlinear systems
- Solve systems with one linear and one quadratic equation
- Solve systems with two quadratic equations
- Apply substitution method to nonlinear systems
- Apply elimination method when appropriate
- Interpret solutions graphically and algebraically
- Solve real-world problems using nonlinear systems
Introduction to Nonlinear Systems
Nonlinear System: A system of equations where at least one equation is not linear. This includes equations with squared terms (x², y²), products of variables (xy), radicals, or other nonlinear relationships.
In previous lessons, we solved linear systems where all equations were lines. Nonlinear systems involve curves such as:
- Parabolas: y = ax² + bx + c
- Circles: (x - h)² + (y - k)² = r²
- Ellipses: (x - h)²/a² + (y - k)²/b² = 1
- Hyperbolas: (x - h)²/a² - (y - k)²/b² = 1
- Other curves: xy = k, y = √x, etc.
Number of Solutions: Unlike linear systems that have 0, 1, or infinitely many solutions, nonlinear systems can have various numbers of solutions:
- No solutions (curves don't intersect)
- One solution (curves are tangent or intersect at one point)
- Two solutions (typical for line and parabola)
- Three solutions (possible with certain curve combinations)
- Four solutions (possible with two circles or conics)
- Infinitely many solutions (curves coincide)
Solution Methods: The two primary algebraic methods are:
- Substitution: Best when one equation is solved for a variable or can be easily solved
- Elimination: Useful when both equations have similar terms (like x² and y²)
Section 1: Systems with One Linear and One Quadratic Equation
When one equation is linear and the other is quadratic, the substitution method is typically most efficient. The linear equation can be solved for one variable and substituted into the quadratic equation.
Example 1: Line and Parabola (Two Solutions)
Solve the system:
y = x² - 4 y = 2x - 1
Step 1: Since both equations are solved for y, set them equal:
x² - 4 = 2x - 1
Step 2: Rearrange to standard form:
x² - 2x - 4 + 1 = 0 x² - 2x - 3 = 0
Step 3: Factor:
(x - 3)(x + 1) = 0
Step 4: Solve for x:
x = 3 or x = -1
Step 5: Find corresponding y-values using y = 2x - 1:
When x = 3: y = 2(3) - 1 = 5 When x = -1: y = 2(-1) - 1 = -3
Solution: (3, 5) and (-1, -3)
Verification: Check both points in both equations.
For (3, 5): 5 = 3² - 4 = 5 and 5 = 2(3) - 1 = 5 For (-1, -3): -3 = (-1)² - 4 = -3 and -3 = 2(-1) - 1 = -3
Example 2: Line and Parabola (One Solution - Tangent)
Solve the system:
y = x² + 2x + 1 y = 2x + 1
Step 1: Set equations equal:
x² + 2x + 1 = 2x + 1
Step 2: Simplify:
x² = 0
Step 3: Solve for x:
x = 0
Step 4: Find y:
y = 2(0) + 1 = 1
Solution: (0, 1)
Interpretation: The line is tangent to the parabola at (0, 1), touching at exactly one point.
Example 3: Line and Parabola (No Solution)
Solve the system:
y = x² + 5 y = x - 2
Step 1: Set equations equal:
x² + 5 = x - 2
Step 2: Rearrange:
x² - x + 7 = 0
Step 3: Check discriminant:
b² - 4ac = (-1)² - 4(1)(7) = 1 - 28 = -27
Conclusion: Since the discriminant is negative, there are no real solutions.
Interpretation: The line and parabola do not intersect.
Example 4: Line and Circle (Two Solutions)
Solve the system:
x² + y² = 25 y = x - 1
Step 1: Substitute y = x - 1 into the circle equation:
x² + (x - 1)² = 25
Step 2: Expand:
x² + x² - 2x + 1 = 25 2x² - 2x + 1 = 25
Step 3: Simplify:
2x² - 2x - 24 = 0 x² - x - 12 = 0
Step 4: Factor:
(x - 4)(x + 3) = 0 x = 4 or x = -3
Step 5: Find y-values:
When x = 4: y = 4 - 1 = 3 When x = -3: y = -3 - 1 = -4
Solutions: (4, 3) and (-3, -4)
Verification:
For (4, 3): 4² + 3² = 16 + 9 = 25 For (-3, -4): (-3)² + (-4)² = 9 + 16 = 25
Example 5: Line and Circle (One Solution - Tangent)
Solve the system:
x² + y² = 16 y = 4
Step 1: Substitute y = 4:
x² + 4² = 16 x² + 16 = 16
Step 2: Solve:
x² = 0 x = 0
Solution: (0, 4)
Interpretation: The horizontal line y = 4 is tangent to the circle at its topmost point.
Example 6: Line and Circle (No Solution)
Solve the system:
x² + y² = 9 y = 5
Step 1: Substitute y = 5:
x² + 5² = 9 x² + 25 = 9 x² = -16
Conclusion: No real solution exists since x² cannot be negative.
Interpretation: The line y = 5 is above the circle (radius 3), so they don't intersect.
Section 2: Systems with Two Quadratic Equations
When both equations are quadratic or involve higher degree terms, we can use either substitution or elimination, depending on the structure of the equations.
Example 7: Two Parabolas
Solve the system:
y = x² - 2 y = -x² + 4
Step 1: Set equations equal:
x² - 2 = -x² + 4
Step 2: Combine like terms:
2x² = 6 x² = 3 x = ±√3
Step 3: Find y-values using y = x² - 2:
When x = √3: y = (√3)² - 2 = 3 - 2 = 1 When x = -√3: y = (-√3)² - 2 = 3 - 2 = 1
Solutions: (√3, 1) and (-√3, 1)
Example 8: Parabola and Circle
Solve the system:
x² + y² = 10 y = x²
Step 1: Substitute y = x² into the circle equation:
x² + (x²)² = 10 x² + x⁴ = 10
Step 2: Rearrange:
x⁴ + x² - 10 = 0
Step 3: Let u = x², then:
u² + u - 10 = 0
Step 4: Use quadratic formula:
u = (-1 ± √(1 + 40))/2 = (-1 ± √41)/2
Step 5: Since u = x², we need u > 0:
u = (-1 + √41)/2 ≈ 2.702 x² ≈ 2.702 x ≈ ±1.644
Step 6: Find y = x²:
y ≈ 2.702
Solutions: Approximately (1.644, 2.702) and (-1.644, 2.702)
Example 9: Two Circles (Two Intersection Points)
Solve the system:
x² + y² = 25 (x - 4)² + y² = 9
Step 1: Expand the second equation:
x² - 8x + 16 + y² = 9 x² + y² - 8x + 16 = 9
Step 2: Substitute x² + y² = 25 from the first equation:
25 - 8x + 16 = 9 41 - 8x = 9 -8x = -32 x = 4
Step 3: Substitute x = 4 into x² + y² = 25:
16 + y² = 25 y² = 9 y = ±3
Solutions: (4, 3) and (4, -3)
Example 10: Hyperbola and Parabola
Solve the system:
xy = 6 y = x² - 5
Step 1: From xy = 6, solve for y:
y = 6/x
Step 2: Set equal to the parabola equation:
6/x = x² - 5
Step 3: Multiply both sides by x (assuming x ≠ 0):
6 = x³ - 5x x³ - 5x - 6 = 0
Step 4: Try x = 2:
2³ - 5(2) - 6 = 8 - 10 - 6 = -8 (not a solution)
Step 5: Try x = 3:
3³ - 5(3) - 6 = 27 - 15 - 6 = 6 (not a solution)
Step 6: Try x = -1:
(-1)³ - 5(-1) - 6 = -1 + 5 - 6 = -2 (not a solution)
Step 7: This cubic requires numerical methods or graphing. Using technology, x ≈ 2.666
y = 6/2.666 ≈ 2.250
Solution: Approximately (2.666, 2.250) plus potentially other solutions from cubic equation.
Example 11: Circle and Ellipse
Solve the system:
x² + y² = 16 x²/9 + y²/4 = 1
Step 1: From the circle equation: y² = 16 - x²
Step 2: Substitute into the ellipse equation:
x²/9 + (16 - x²)/4 = 1
Step 3: Multiply by 36 to clear denominators:
4x² + 9(16 - x²) = 36 4x² + 144 - 9x² = 36 -5x² = -108 x² = 21.6
Step 4: This gives x² = 21.6, but x² must be ≤ 16 for the circle.
Conclusion: No real solution exists. The circle and ellipse do not intersect.
Section 3: Solving by Substitution
The substitution method works by solving one equation for a variable and substituting into the other. This is particularly effective when one equation is already solved for a variable or can be easily manipulated.
Example 12: Circle and Line Using Substitution
Solve the system:
(x - 2)² + (y + 1)² = 10 x + 2y = 3
Step 1: Solve the linear equation for x:
x = 3 - 2y
Step 2: Substitute into the circle equation:
((3 - 2y) - 2)² + (y + 1)² = 10 (1 - 2y)² + (y + 1)² = 10
Step 3: Expand:
1 - 4y + 4y² + y² + 2y + 1 = 10 5y² - 2y + 2 = 10
Step 4: Simplify:
5y² - 2y - 8 = 0
Step 5: Use quadratic formula:
y = (2 ± √(4 + 160))/10 = (2 ± √164)/10 = (2 ± 2√41)/10 = (1 ± √41)/5
Step 6: Calculate approximate values:
y₁ = (1 + √41)/5 ≈ 1.481 y₂ = (1 - √41)/5 ≈ -1.081
Step 7: Find corresponding x-values:
x₁ = 3 - 2(1.481) ≈ 0.038 x₂ = 3 - 2(-1.081) ≈ 5.162
Solutions: Approximately (0.038, 1.481) and (5.162, -1.081)
Example 13: Ellipse and Line
Solve the system:
x²/16 + y²/9 = 1 2x - y = 4
Step 1: Solve for y:
y = 2x - 4
Step 2: Substitute into ellipse equation:
x²/16 + (2x - 4)²/9 = 1
Step 3: Multiply by 144:
9x² + 16(2x - 4)² = 144 9x² + 16(4x² - 16x + 16) = 144 9x² + 64x² - 256x + 256 = 144
Step 4: Simplify:
73x² - 256x + 112 = 0
Step 5: Use quadratic formula:
x = (256 ± √(65536 - 32704))/146 = (256 ± √32832)/146 = (256 ± 181.19)/146
Step 6: Calculate:
x₁ ≈ (256 + 181.19)/146 ≈ 2.994 x₂ ≈ (256 - 181.19)/146 ≈ 0.512
Step 7: Find y-values:
y₁ = 2(2.994) - 4 ≈ 1.988 y₂ = 2(0.512) - 4 ≈ -2.976
Solutions: Approximately (2.994, 1.988) and (0.512, -2.976)
Example 14: Two Conics
Solve the system:
x² - y² = 7 x² + y = 13
Step 1: Solve the second equation for x²:
x² = 13 - y
Step 2: Substitute into the first equation:
(13 - y) - y² = 7 -y² - y + 13 = 7 -y² - y + 6 = 0 y² + y - 6 = 0
Step 3: Factor:
(y + 3)(y - 2) = 0 y = -3 or y = 2
Step 4: Find x-values:
When y = -3: x² = 13 - (-3) = 16, so x = ±4 When y = 2: x² = 13 - 2 = 11, so x = ±√11
Solutions: (4, -3), (-4, -3), (√11, 2), (-√11, 2)
Example 15: System Requiring Quadratic Formula
Solve the system:
x² + y² = 13 x - y = 1
Step 1: Solve for x:
x = y + 1
Step 2: Substitute:
(y + 1)² + y² = 13 y² + 2y + 1 + y² = 13 2y² + 2y - 12 = 0 y² + y - 6 = 0
Step 3: Factor:
(y + 3)(y - 2) = 0 y = -3 or y = 2
Step 4: Find x-values:
When y = -3: x = -3 + 1 = -2 When y = 2: x = 2 + 1 = 3
Solutions: (-2, -3) and (3, 2)
Example 16: System with Radical Expressions
Solve the system:
y = √x x² + y² = 20
Step 1: Substitute y = √x:
x² + (√x)² = 20 x² + x = 20 x² + x - 20 = 0
Step 2: Factor:
(x + 5)(x - 4) = 0 x = -5 or x = 4
Step 3: Check validity (x must be ≥ 0 for √x):
x = -5 is invalid (negative) x = 4 is valid
Step 4: Find y:
y = √4 = 2
Solution: (4, 2)
Section 4: Solving by Elimination
Elimination works well when both equations have similar terms, particularly x² and y² terms with the same or easily managed coefficients. We add or subtract equations to eliminate one variable.
Example 17: Eliminating x² and y² Terms
Solve the system:
x² + y² = 25 x² - y² = 7
Step 1: Add the equations to eliminate y²:
x² + y² = 25 + x² - y² = 7 _______________ 2x² = 32
Step 2: Solve for x:
x² = 16 x = ±4
Step 3: Substitute into first equation:
16 + y² = 25 y² = 9 y = ±3
Solutions: (4, 3), (4, -3), (-4, 3), (-4, -3)
Example 18: Eliminating y² Terms
Solve the system:
2x² + 3y² = 21 x² + 3y² = 12
Step 1: Subtract the second equation from the first:
2x² + 3y² = 21 - (x² + 3y²) = 12 _________________ x² = 9
Step 2: Solve for x:
x = ±3
Step 3: Substitute into second equation:
9 + 3y² = 12 3y² = 3 y² = 1 y = ±1
Solutions: (3, 1), (3, -1), (-3, 1), (-3, -1)
Example 19: Eliminating with Different Coefficients
Solve the system:
3x² + 2y² = 35 x² + y² = 13
Step 1: Multiply second equation by -2:
3x² + 2y² = 35 -2x² - 2y² = -26
Step 2: Add equations:
3x² + 2y² = 35 + (-2x² - 2y²) = -26 ___________________ x² = 9
Step 3: Solve for x:
x = ±3
Step 4: Substitute into x² + y² = 13:
9 + y² = 13 y² = 4 y = ±2
Solutions: (3, 2), (3, -2), (-3, 2), (-3, -2)
Example 20: Circle and Ellipse Problem
Solve the system:
x² + y² = 20 4x² + 9y² = 80
Step 1: Multiply first equation by -4:
-4x² - 4y² = -80 4x² + 9y² = 80
Step 2: Add equations:
5y² = 0 y² = 0 y = 0
Step 3: Substitute into first equation:
x² + 0 = 20 x² = 20 x = ±2√5
Solutions: (2√5, 0) and (-2√5, 0)
Section 5: Applications of Nonlinear Systems
Nonlinear systems appear in many real-world contexts, from geometry to physics to business applications.
Example 21: Rectangle Geometry Problem
A rectangle has a perimeter of 36 meters and an area of 80 square meters. Find its dimensions.
Step 1: Set up variables:
Let l = length and w = width
Step 2: Write equations:
Perimeter: 2l + 2w = 36 → l + w = 18 Area: lw = 80
Step 3: Solve for l:
l = 18 - w
Step 4: Substitute into area equation:
(18 - w)w = 80 18w - w² = 80 -w² + 18w - 80 = 0 w² - 18w + 80 = 0
Step 5: Factor:
(w - 10)(w - 8) = 0 w = 10 or w = 8
Step 6: Find corresponding lengths:
When w = 10: l = 18 - 10 = 8 When w = 8: l = 18 - 8 = 10
Answer: The dimensions are 10 m by 8 m (or equivalently 8 m by 10 m).
Example 22: Projectile and Target
A projectile follows the path y = -x² + 4x, where x and y are measured in meters. A circular target is positioned with equation x² + y² = 16. Where does the projectile hit the target?
Step 1: Set up the system:
y = -x² + 4x x² + y² = 16
Step 2: Substitute first equation into second:
x² + (-x² + 4x)² = 16 x² + x⁴ - 8x³ + 16x² = 16 x⁴ - 8x³ + 17x² - 16 = 0
Step 3: This requires numerical methods. Using technology, one positive solution is x ≈ 1.464
Step 4: Find y:
y = -(1.464)² + 4(1.464) ≈ 3.712
Answer: The projectile hits the target at approximately (1.464, 3.712) meters.
Example 23: Business - Cost and Revenue
A company's total cost is C = x² + 10x + 100 and total revenue is R = 50x, where x is the number of units produced (in hundreds). At what production levels does the company break even (cost equals revenue)?
Step 1: Set cost equal to revenue:
x² + 10x + 100 = 50x
Step 2: Rearrange:
x² - 40x + 100 = 0
Step 3: Use quadratic formula:
x = (40 ± √(1600 - 400))/2 = (40 ± √1200)/2 = (40 ± 20√3)/2 = 20 ± 10√3
Step 4: Calculate:
x₁ = 20 + 10√3 ≈ 37.32 (hundreds of units) x₂ = 20 - 10√3 ≈ 2.68 (hundreds of units)
Answer: The company breaks even at approximately 268 units and 3,732 units.
Example 24: Optimization with Constraint
Find two positive numbers whose sum is 20 and whose product is maximum.
Step 1: Set up variables and equations:
Let x and y be the two numbers Constraint: x + y = 20 Maximize: P = xy
Step 2: Express y in terms of x:
y = 20 - x
Step 3: Write product as function of one variable:
P = x(20 - x) = 20x - x²
Step 4: This is a parabola opening downward. Maximum occurs at vertex.
x = -b/(2a) = -20/(2(-1)) = 10
Step 5: Find y:
y = 20 - 10 = 10
Step 6: Maximum product:
P = 10 × 10 = 100
Answer: The two numbers are both 10, giving a maximum product of 100.
Example 25: Number Problem
The sum of two numbers is 12, and the sum of their squares is 74. Find the numbers.
Step 1: Set up system:
x + y = 12 x² + y² = 74
Step 2: Solve first equation for y:
y = 12 - x
Step 3: Substitute:
x² + (12 - x)² = 74 x² + 144 - 24x + x² = 74 2x² - 24x + 144 = 74 2x² - 24x + 70 = 0 x² - 12x + 35 = 0
Step 4: Factor:
(x - 7)(x - 5) = 0 x = 7 or x = 5
Step 5: Find corresponding y-values:
When x = 7: y = 12 - 7 = 5 When x = 5: y = 12 - 5 = 7
Answer: The two numbers are 7 and 5.
Example 26: Garden Design
A landscaper is designing a rectangular garden with a diagonal walkway. The garden is 4 meters longer than it is wide, and the diagonal walkway is 20 meters long. Find the dimensions of the garden.
Step 1: Set up variables:
Let w = width Then length = w + 4
Step 2: Use Pythagorean theorem:
w² + (w + 4)² = 20² w² + w² + 8w + 16 = 400 2w² + 8w + 16 = 400 2w² + 8w - 384 = 0 w² + 4w - 192 = 0
Step 3: Use quadratic formula:
w = (-4 ± √(16 + 768))/2 = (-4 ± √784)/2 = (-4 ± 28)/2
Step 4: Calculate:
w = (-4 + 28)/2 = 12 (positive solution) w = (-4 - 28)/2 = -16 (reject negative)
Step 5: Find length:
length = 12 + 4 = 16 meters
Answer: The garden is 12 meters wide and 16 meters long.
Verification: 12² + 16² = 144 + 256 = 400 = 20²
Check Your Understanding
1. How many solutions can a system consisting of a line and a circle have? Explain each case.
A line and circle can have:
- 0 solutions: The line doesn't intersect the circle (line is too far away)
- 1 solution: The line is tangent to the circle (touches at exactly one point)
- 2 solutions: The line crosses through the circle (enters and exits)
2. Solve the system: y = x² + 1 and y = 2x + 4
Set equations equal: x² + 1 = 2x + 4
Rearrange: x² - 2x - 3 = 0
Factor: (x - 3)(x + 1) = 0
x = 3 or x = -1
When x = 3: y = 2(3) + 4 = 10
When x = -1: y = 2(-1) + 4 = 2
Solutions: (3, 10) and (-1, 2)
3. Solve the system: x² + y² = 25 and y = x + 1
Substitute y = x + 1 into circle equation:
x² + (x + 1)² = 25
x² + x² + 2x + 1 = 25
2x² + 2x - 24 = 0
x² + x - 12 = 0
(x + 4)(x - 3) = 0
x = -4 or x = 3
When x = -4: y = -4 + 1 = -3
When x = 3: y = 3 + 1 = 4
Solutions: (-4, -3) and (3, 4)
4. Solve the system: x² + y² = 10 and x² - y² = 2
Add the equations:
2x² = 12, so x² = 6, x = ±√6
Substitute into first equation: 6 + y² = 10
y² = 4, so y = ±2
Solutions: (√6, 2), (√6, -2), (-√6, 2), (-√6, -2)
5. Which method (substitution or elimination) would you use for the system: 3x² + 4y² = 48 and x² + 4y² = 24? Why?
Elimination method is best because both equations have the term 4y², making it easy to eliminate.
Subtract the second equation from the first:
3x² + 4y² - (x² + 4y²) = 48 - 24
2x² = 24
This immediately gives x² = 12, making the solution straightforward.
6. Solve the system: xy = 12 and x + y = 7
From second equation: y = 7 - x
Substitute into first: x(7 - x) = 12
7x - x² = 12
-x² + 7x - 12 = 0
x² - 7x + 12 = 0
(x - 3)(x - 4) = 0
x = 3 or x = 4
When x = 3: y = 7 - 3 = 4
When x = 4: y = 7 - 4 = 3
Solutions: (3, 4) and (4, 3)
7. A rectangle has a perimeter of 28 cm and an area of 45 cm². Find its dimensions.
Let l = length, w = width
Perimeter: 2l + 2w = 28, so l + w = 14
Area: lw = 45
From first equation: l = 14 - w
Substitute: (14 - w)w = 45
14w - w² = 45
w² - 14w + 45 = 0
(w - 9)(w - 5) = 0
w = 9 or w = 5
Corresponding lengths: l = 5 or l = 9
Answer: 9 cm by 5 cm
8. The difference of two numbers is 3, and the sum of their squares is 89. Find the numbers.
Let x and y be the numbers
x - y = 3
x² + y² = 89
From first equation: x = y + 3
Substitute: (y + 3)² + y² = 89
y² + 6y + 9 + y² = 89
2y² + 6y - 80 = 0
y² + 3y - 40 = 0
(y + 8)(y - 5) = 0
y = -8 or y = 5
When y = 5: x = 5 + 3 = 8
When y = -8: x = -8 + 3 = -5
Numbers: 8 and 5 (or -5 and -8)
Common Mistakes to Avoid:
- Forgetting to check if solutions satisfy the domain restrictions (e.g., x ≥ 0 for √x)
- Missing solutions by only considering positive square roots
- Not verifying solutions in both original equations
- Arithmetic errors when expanding squared binomials
- Choosing an inefficient method (use elimination when terms align nicely)
Key Strategies:
- Visualize the graphs mentally to predict the number of solutions
- Use substitution when one equation is linear or easily solved
- Use elimination when both equations have x² and y² terms
- Always check your solutions in both original equations
- Consider the domain and range when interpreting solutions
- For application problems, check that solutions make sense in context