Lesson 112: Graphing and Solving Systems of Linear and Quadratic Equations

New Concept

A system of equations can also consist of a linear equation and a quadratic equation.

What’s next

Next, you’ll master key methods for solving these systems: graphing to visualize intersections and substitution to find exact algebraic solutions.

Property

A system of one linear and one quadratic equation can have zero, one, or two solutions.

Explanation

Imagine a line meeting a parabola. They can cross twice (two solutions), touch at one point (one solution), or miss each other completely (no solution). The number of intersections is the number of solutions you have!

Examples

Two solutions: $y = x^2$ and $y = 4$. One solution: $y = x^2$ and $y = 0$. No solution: $y = x^2$ and $y = -4$.

Property

To solve a system of equations by graphing, graph both equations on the same coordinate plane. The solution is the set of ordered pairs of the intersection points.

Explanation

Graphing is a visual treasure hunt! Just draw the line and the parabola. The exact spots where they cross are your solutions—the $(x, y)$ pairs that work for both equations.

Examples

For $y = x^2$ and $y = x+2$, the graphs intersect at the points $(-1, 1)$ and $(2, 4)$, which are the two solutions to the system.

Let's see where a parabola and a line meet by drawing their paths. The first key idea is solving by graphing, which allows us to visualize the solutions.

Example Problem
Solve the system by graphing: $y = 3x^2 - 5$ and $y = 6x - 5$.

Step-by-Step

1. Graph the parabola $y = 3x^2 - 5$ and the line $y = 6x - 5$ on the same coordinate plane.
2. The graphs show two points where the parabola and line intersect.
3. By observing the graph, we can identify the coordinates of these two points as $(0, -5)$ and $(2, 7)$.
4. Now, we must check the first solution, $(0, -5)$, in both original equations to verify it.
5. In $y = 3x^2 - 5$, we check: $-5 \stackrel{?}{=} 3(0)^2 - 5$, which simplifies to $-5 = -5$. This is correct.
6. In $y = 6x - 5$, we check: $-5 \stackrel{?}{=} 6(0) - 5$, which simplifies to $-5 = -5$. This is also correct.
7. Next, let's verify the second solution, $(2, 7)$.
8. In $y = 3x^2 - 5$, we check: $7 \stackrel{?}{=} 3(2)^2 - 5$, which simplifies to $7 = 3(4) - 5$, or $7 = 12 - 5$. This is correct.
9. In $y = 6x - 5$, we check: $7 \stackrel{?}{=} 6(2) - 5$, which simplifies to $7 = 12 - 5$. This is also correct.

Property

Set the expressions for $y$ from each equation equal to each other. Solve the new equation for all possible $x$-values, then find their matching $y$-values.

Explanation

If both equations equal $y$, they must equal each other! Set them together to make one equation. Solve for $x$, then plug that value back into an original equation to find its $y$ partner.

Examples

System: $y = x^2 - x$ and $y = 2x+4$. Set $x^2 - x = 2x+4$, which becomes $x^2 - 3x - 4 = 0$.

Sometimes graphing isn't precise. Let's use algebra to find the exact solutions. The second key idea is solving using substitution, which gives us precise coordinates.

Example Problem
Solve the system using substitution: $y = x^2 + 3x - 2$ and $y = 3x + 7$.

Step-by-Step

1. Since both equations are equal to $y$, we can set their right sides equal to each other. This is the substitution step.

2. To solve for $x$, we need to get the equation into the standard quadratic form ($ax^2+bx+c=0$). We can do this by adding the expression $-3x - 7$ to both sides.

3. We should recognize the left side of the equation, $x^2 - 9$, as a difference of two squares.

4. Now, we use the Zero Product Property and set each factor to zero to find the two possible values for $x$.

5. We have two $x$-values, so we must find the corresponding $y$-value for each. We can use the simpler linear equation, $y = 3x + 7$.
6. First, substitute $x = -3$:

7. Next, substitute $x = 3$:

8. The solutions are the ordered pairs $(-3, -2)$ and $(3, 16)$.