Learn on PengienVision, Algebra 2Chapter 2: Quadratic Functions and Equations

Lesson 7: Linear-Quadratic Systems

In this Grade 11 enVision Algebra 2 lesson, students learn to solve linear-quadratic systems by analyzing how a line and a parabola can intersect in 0, 1, or 2 points. They apply the substitution method to find exact solution pairs and use factoring to solve the resulting quadratic equations. Real-world applications, such as modeling a ball's parabolic path against a linear hill, reinforce how these algebraic techniques translate to practical problem-solving.

Section 1

Solving systems by graphing

Property

Every point on the graph of an equation represents a solution to that equation. A solution to a system of two equations must be a point that lies on both graphs. Therefore, a solution to the system is a point where the two graphs intersect.

Examples

For the system $y = x + 2$ and $y = -x + 4$ , graphing the lines shows they intersect at the point $(1, 3)$ . Therefore, the solution to the system is $x=1, y=3$ .

To solve the system $y = 1.7x + 0.4$ and $y = 4.1x + 5.2$ , we can graph both lines. The intersection point occurs at $(-2, -3)$ , which is the solution to the system.

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 $y$ , we can set the expressions equal to each other. This creates a single equation in terms of $x$ , 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 $y = x^2 - 2x + 3$ and $y = x + 1$ , set the expressions for $y$ equal: $x^2 - 2x + 3 = x + 1$ . This simplifies to $x^2 - 3x + 2 = 0$ , which factors into $(x-1)(x-2)=0$ . The solutions are $x=1$ and $x=2$ . The corresponding points are $(1, 2)$ and $(2, 3)$ .

A company's cost is $C = 0.1x^2 + 2x + 100$ and revenue is $R = 12x$ . To break even, $C=R$ . So, $0.1x^2 + 2x + 100 = 12x$ . This gives $0.1x^2 - 10x + 100 = 0$ . Using the quadratic formula, break-even points are at $x=11.27$ and $x=88.73$ items.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Solving systems by graphing

Property

Examples

For the system $y = x + 2$ and $y = -x + 4$ , graphing the lines shows they intersect at the point $(1, 3)$ . Therefore, the solution to the system is $x=1, y=3$ .

To solve the system $y = 1.7x + 0.4$ and $y = 4.1x + 5.2$ , we can graph both lines. The intersection point occurs at $(-2, -3)$ , which is the solution to the system.

Section 2

Systems with a Quadratic Equation

Property

Examples

To solve the system $y = x^2 - 2x + 3$ and $y = x + 1$ , set the expressions for $y$ equal: $x^2 - 2x + 3 = x + 1$ . This simplifies to $x^2 - 3x + 2 = 0$ , which factors into $(x-1)(x-2)=0$ . The solutions are $x=1$ and $x=2$ . The corresponding points are $(1, 2)$ and $(2, 3)$ .

A company's cost is $C = 0.1x^2 + 2x + 100$ and revenue is $R = 12x$ . To break even, $C=R$ . So, $0.1x^2 + 2x + 100 = 12x$ . This gives $0.1x^2 - 10x + 100 = 0$ . Using the quadratic formula, break-even points are at $x=11.27$ and $x=88.73$ items.

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Solving systems by graphing

Property

Examples

For the system $y = x + 2$ and $y = -x + 4$ , graphing the lines shows they intersect at the point $(1, 3)$ . Therefore, the solution to the system is $x=1, y=3$ .

To solve the system $y = 1.7x + 0.4$ and $y = 4.1x + 5.2$ , we can graph both lines. The intersection point occurs at $(-2, -3)$ , which is the solution to the system.

Section 2

Systems with a Quadratic Equation

Property

Examples

To solve the system $y = x^2 - 2x + 3$ and $y = x + 1$ , set the expressions for $y$ equal: $x^2 - 2x + 3 = x + 1$ . This simplifies to $x^2 - 3x + 2 = 0$ , which factors into $(x-1)(x-2)=0$ . The solutions are $x=1$ and $x=2$ . The corresponding points are $(1, 2)$ and $(2, 3)$ .

A company's cost is $C = 0.1x^2 + 2x + 100$ and revenue is $R = 12x$ . To break even, $C=R$ . So, $0.1x^2 + 2x + 100 = 12x$ . This gives $0.1x^2 - 10x + 100 = 0$ . Using the quadratic formula, break-even points are at $x=11.27$ and $x=88.73$ items.