Solving Equations Page 3

Page 3

Solving Linear-Quadratic Systems of Equations

You have learned how to solve linear pairs and linear systems of equations in the past.
You can take a quick review of solving linear systems of equations if you wish.
Now you will look at solving linear-quadratic systems of equations.
A linear - quadratic system of equations contains a linear equation and a quadratic equation. Solving the system means finding pairs of real numbers x and y that simultaneously satisfy both equations. That is, you find the points of intersection between the graphs of the equations.
By graphing both equations, you can get an idea of how many solutions there are and an estimate of their value. By algebraic manipulations, you can find the exact values.
Let's look at an example to see how to solve these types of systems.

EXAMPLE 1:

Solve the following linear-quadratic system by graphing:
1. y = x + 1
2. x² + y² = 25

The equation y = x + 1 defines a line with y-intercept at (0, 1) and a slope of 1.
The equation x² + y² = 25 defines a circle centred at the origin with a radius of 5 units.

You can sketch a graph of both equations to get an idea of where they intersect. This is done for you in Figure 1. FIGURE 1

Notice that the circle and line intersect at 2 places which appear to be (3, 4) and (-4, -3). You can now easily check that x = 3 and y = 4 is a solution to both y = x + 1 and x² + y² = 25. Likewise for x = -4 and y = -3.

EXAMPLE 1:
Solve the following linear-quadratic system by graphing: 1. y = x + 1 2. x² + y² = 25
The equation y = x + 1 defines a line with y-intercept at (0, 1) and a slope of 1. The equation x² + y² = 25 defines a circle centred at the origin with a radius of 5 units.
You can sketch a graph of both equations to get an idea of where they intersect. This is done for you in Figure 1.	FIGURE 1
Notice that the circle and line intersect at 2 places which appear to be (3, 4) and (-4, -3). You can now easily check that x = 3 and y = 4 is a solution to both y = x + 1 and x² + y² = 25. Likewise for x = -4 and y = -3.

How can you find exact solutions for a pair of equations when intersection points of the graphs are not so easy to see? The answer is to use algebra and maybe use the graphs to give you confidence in your calculations.
Just as with linear systems of equations, there are several algebraic strategies which can be used to solve linear-quadratic systems of equations.

Some of the common techniques are given names - comparison, elimination and substitution.
Each situation calls for its own stratetgy and each person prefers a different method. It is up to you to choose which algebraic method, or combination of methods you use for each problem.
In Example 1, a good strategy (but not the only stategy) for solving the system would be the substitution method:

Notice that the system of equations in Example 1 had 2 solutions. Some linear-quadratic systems may have only 1 solution, while other linear-quadratic systems may have no solutions.
When a system has no solutions, the graphs of the equations do not intersect and no point will satisfy both equations.
Some great practice questions.