An equation in which one or more terms have a variable of degree 2 or higher is called a nonlinear equation. A nonlinear system of equations contains at least one nonlinear equation.

When solving a system in which one equation is linear, it is usually easiest to use the substitution method. Solve the linear equation for either x or y, then substitute the resulting expression into the nonlinear equation.

Example. Solve the system:

2x + 3y = 12 Equation 1

Equation 2

Solution.

2x + 3y = 12 2x = 12 – 3y x = 6 -	Solve Equation 1, the linear equation, for y
144 – 72y + 9y + 9y = 144 18y - 72y = 0	Substitute the expression obtained in step 1, 6 - , in place of x in Equation 2; simplify; solve for y
18y(y – 4) = 0 18y = 0 or y – 4 = 0 y = 0 or y = 4	Solve by factoring
x = 6 - x = 6 - x = 6 So, (6,0) is one solution	Back-substitute each value of y into the expression obtained in step 1: x = 6 -
x = 6 - x = 6 - x = 0 So, (0,4) is another solution.

The addition method works well on nonlinear systems when each equation is in the form

. If necessary, we multiply either equation or both equations by numbers so that the coefficients of x or y will have the sum of 0. We then add the equations. The sum will be an equation in one variable.

Example. Solve:

Equation 1

Equation 2

Solution. We will multiply both sides of equation 1 by 3 and both sides of equation 2 by 4.

3( Equation 3

4( Equation 4

In equations 3 and 4 obtained above, the coefficients of y are 12 and –12; when we add the equations together, y will be eliminated.

Equation 3

Equation 4

17x = 68

x = 4

x = ± 2

Now we back-substitute each of these values of x into either one of the original equations:

x = 2 y = ± 1 This means that when x = 2, y can be either 1 or –1. So there are two solutions: (2,1) and (2,-1).

x = -2 y = ± 1 When x = -2, y can be either 1 or -1, giving us two additional solutions: (-2, 1) and (-2, -1).

Example. Solve:

xy = 4 Equation 1

Equation 2

Solution. We will solve this system using the substitution method, and begin by solving Equation 1 for y:

xy = 4

y =

Substitute for y in Equation 2:

	Equation 2
	Substitute for y
	Simplify
	Multiply both sides of the equation by x to eliminate the fraction
2x + 16 = 18x	Distributive Property
2x - 18x+ 16 = 0	Solve by factoring
x - 9x+ 8 = 0	Divide both sides by 2
(x - 8)(x - 1) = 0 x - 8 = 0 or x - 1 = 0 x = 8 or x = 1 x = or x = x = or x =

Each of these values of x will be back-substituted into the expression obtained in step 1,

y = .

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