Systems of Non-Linear Equations

Learning Objective

• Solve systems of equations involving linear, quadratic, and other non-linear functions.

A system of equationsA set of two or more equations that share two or more unknowns. is a set of two or more equations that share two or more unknowns. The solutions to a system of equations are all the values that make all of the equations true, or the points where the graphs of the equations intersect.

We can solve a system of linear equations through graphing, substitution, and linear combination. Systems of nonlinear functionsA function with a variable rate of change that graphs as a curved line., such as quadratic or exponential equations, can be handled with the same techniques.

To illustrate how to solve these systems, we'll focus on linear and quadratic systems with just two equations. But be aware that systems can be larger and more complex than these examples.

Let’s begin by talking about two linear equations. The solution to this kind of system is the point of intersection between the two lines, or the place where the two equations have the same `x` and `y` values. There may be one solution, no solution, or an infinite number of solutions to a system of two linear equations:

To solve a system of a linear equation and a quadratic equation, we will do the same thing, find the point—or points—of intersection between the two graphs:

It does not make sense to consider the case where the two equations represent the same set of points, since a straight line can never be a parabola, and vice versa.

Notice that this means the possible number of solutions for a system of two linear equations is `0` (never intersect), `1` (intersect in one place), or infinity (the lines are identical). The number of solutions for a system with one linear equation and one quadratic equation is either `0` (never intersect), `1` (intersect in one place), or `2` (intersect in two places).

Let's solve a system of a linear and a quadratic equation by graphing:

Notice that while we can tell approximately where the two graphs intersect, it is difficult to find an exact location.

Now let’s solve the same system using substitution. When solving using substitution, use the following steps:

`1`. Select one equation and isolate a variable. (Choose an equation and variable that are easy to isolate!)

`2`. Substitute the resulting expression for a variable into the other equation, every time that variable appears.

`3`. Solve the second equation for the second variable.

`4`. Substitute the solution to step `3` into the expression in step `1`, to find the other variable.

Example
Problem	Solve the system using the substitution method `y=x^2-5` and `y=3x+7`.
`x^2-5=3x+7`				In this case, both equations have `y` isolated, so we can set them equal.
`x^2-3x-12=0`				Subtract `3x` from both sides and subtract `7` from both sides. Now there is a quadratic equation set equal to `0` and we can use the quadratic formulaThe formula `x=(-b+-sqrt(b^2-4ac))/(2a)`; it is used to solve a quadratic equation of the form `ax^2+bx+c=0`., `x=(-b+-sqrt(b^2-4ac))/(2a)`, to find the solution.
`x^2` `-3x` `-12` `=` `0`     `a=1` `b=-3` `c=-12`				`a=1`, `b=-3`, and `c = -12`.
`x=(-(-3)+-sqrt((-3)^2-4(1)(-12)))/(2(1))`				Substitute the values of `a`, `b`, and `c` into the formula.
`x=(3+-sqrt(9+48))/2=(3+-sqrt57)/2`				Simplify.
`x~~5.27` or `x~~-2.27`				Simplify some more, remembering to evaluate both `(3+sqrt57)/2` and `(3-sqrt57)/2`.
	`y=3x+7`   `y=3(5.27)+7`	`y=3x+7`   `y=3(-2.27)+7`		Evaluate either function for each `x` to find the corresponding `y` value.
Answer	`(5.27, 22.82)` and `(-2.27, 0.18)`

By using substitution, we have arrived at a more precise answer than we did when graphing this system. However, approximations were made when we took the square root of `57`. This still is not an exact solution!

As the algebra of substitution can become quite involved, it is always a good idea to check the result in the original equations.

Here is a check of the point `(5.27, 22.82)`:

Notice that the check doesn’t result in perfect equality, but it's close.

Now let’s look at two quadratic equations. Picture for a moment how the graphs of two quadratic equations can intersect (or not).

We can solve a system of quadratic equations by graphing:

Again, we can't be certain by graphing that our solutions are exact. An algebraic method can ensure an exact solution every time. For example, we can solve a system of quadratic equations by substitution:

Since there is no solution, we cannot check our answer algebraically, but we can look at the graphs to verify that there is no solution:

We can also use linear combination to solve systems of equations, following these steps:

`1`. Rearrange equations so the terms line up.

`2`. Multiply none, or one, or both equations by a constant so that the coefficients of one of the variables are opposites.

`3`. Add the equations to eliminate one of the variables.

`4`. Solve the resulting equation.

`5`. Substitute the solution in step `4` back into an original equation to find the other variable.

Ready to give that a try?

Solving systems of non-linear equations can be done using the techniques of graphing, substitution, and linear combination. Just like with systems of linear equations, when we find solutions to systems of non-linear equations, we are finding the intersection of their graphs or the places where the equations have the same variable values.