Learn on PengiBig Ideas Math, Algebra 2Chapter 3: Quadratic Equations and Complex Numbers

Lesson 5: Solving Nonlinear Systems

In this Grade 8 lesson from Big Ideas Math Algebra 2, Chapter 3, students learn how to solve systems of nonlinear equations using graphing, substitution, and elimination methods. The lesson covers systems that pair a quadratic equation with a linear equation or two quadratic equations, exploring how these graphs can intersect in zero, one, or two points. Students practice identifying solutions analytically and graphically, building on their prior knowledge of linear systems.

Section 1

System of Nonlinear Equations

Property

A system of nonlinear equations is a system where at least one of the equations is not linear.

Just as with systems of linear equations, a solution of a nonlinear system is an ordered pair that makes both equations true. In a nonlinear system, there may be more than one solution. The graphs may be circles, parabolas or hyperbolas and there may be several points of intersection, and so several solutions.

Examples

  • A system with a parabola and a circle:
{y=x21x2+y2=9\begin{cases} y = x^2 - 1 \\ x^2 + y^2 = 9 \end{cases}

Section 2

Systems with a Quadratic Equation

Property

The solution to a system of equations is the set of intersection points of their graphs. For a system with a quadratic equation, we can solve it algebraically. If both equations are solved for yy, we can set the expressions equal to each other. This creates a single equation in terms of xx, which can then be solved. The resulting equation is often quadratic and can be solved using methods like the quadratic formula.

Examples

  • To solve the system y=x22x+3y = x^2 - 2x + 3 and y=x+1y = x + 1, set the expressions for yy equal: x22x+3=x+1x^2 - 2x + 3 = x + 1. This simplifies to x23x+2=0x^2 - 3x + 2 = 0, which factors into (x1)(x2)=0(x-1)(x-2)=0. The solutions are x=1x=1 and x=2x=2. The corresponding points are (1,2)(1, 2) and (2,3)(2, 3).
  • A company's cost is C=0.1x2+2x+100C = 0.1x^2 + 2x + 100 and revenue is R=12xR = 12x. To break even, C=RC=R. So, 0.1x2+2x+100=12x0.1x^2 + 2x + 100 = 12x. This gives 0.1x210x+100=00.1x^2 - 10x + 100 = 0. Using the quadratic formula, break-even points are at x=11.27x=11.27 and x=88.73x=88.73 items.

Book overview

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Chapter 3: Quadratic Equations and Complex Numbers

  1. Lesson 1

    Lesson 1: Solving Quadratic Equations

  2. Lesson 2

    Lesson 3: Completing the Square

  3. Lesson 3

    Lesson 4: Using the Quadratic Formula

  4. Lesson 4Current

    Lesson 5: Solving Nonlinear Systems

  5. Lesson 5

    Lesson 6: Quadratic Inequalities

Lesson overview

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Section 1

System of Nonlinear Equations

Property

A system of nonlinear equations is a system where at least one of the equations is not linear.

Just as with systems of linear equations, a solution of a nonlinear system is an ordered pair that makes both equations true. In a nonlinear system, there may be more than one solution. The graphs may be circles, parabolas or hyperbolas and there may be several points of intersection, and so several solutions.

Examples

  • A system with a parabola and a circle:
{y=x21x2+y2=9\begin{cases} y = x^2 - 1 \\ x^2 + y^2 = 9 \end{cases}

Section 2

Systems with a Quadratic Equation

Property

The solution to a system of equations is the set of intersection points of their graphs. For a system with a quadratic equation, we can solve it algebraically. If both equations are solved for yy, we can set the expressions equal to each other. This creates a single equation in terms of xx, which can then be solved. The resulting equation is often quadratic and can be solved using methods like the quadratic formula.

Examples

  • To solve the system y=x22x+3y = x^2 - 2x + 3 and y=x+1y = x + 1, set the expressions for yy equal: x22x+3=x+1x^2 - 2x + 3 = x + 1. This simplifies to x23x+2=0x^2 - 3x + 2 = 0, which factors into (x1)(x2)=0(x-1)(x-2)=0. The solutions are x=1x=1 and x=2x=2. The corresponding points are (1,2)(1, 2) and (2,3)(2, 3).
  • A company's cost is C=0.1x2+2x+100C = 0.1x^2 + 2x + 100 and revenue is R=12xR = 12x. To break even, C=RC=R. So, 0.1x2+2x+100=12x0.1x^2 + 2x + 100 = 12x. This gives 0.1x210x+100=00.1x^2 - 10x + 100 = 0. Using the quadratic formula, break-even points are at x=11.27x=11.27 and x=88.73x=88.73 items.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 3: Quadratic Equations and Complex Numbers

  1. Lesson 1

    Lesson 1: Solving Quadratic Equations

  2. Lesson 2

    Lesson 3: Completing the Square

  3. Lesson 3

    Lesson 4: Using the Quadratic Formula

  4. Lesson 4Current

    Lesson 5: Solving Nonlinear Systems

  5. Lesson 5

    Lesson 6: Quadratic Inequalities