# Algebra Tips: Solving 2 Non-Linear Equations in 2 Variables

A set of non-linear equations in 2 variables.

Systems of equations in several variables are also known as simultaneous equations. The simplest example of this kind of algebra problem is when you have two liinear equations in two variables, such as

2x + 5y = 7
3x - 8y = 1

The equations do not necessarily have to be linear, though most textbook examples begin with linear systems since they are easier to solve. Since there are thousands of tutorials online for solving linear systems, this article will show you how to solve non-linear systems of equations using the substitution method.

## Example 1, Step 1

Select one equation and isolate one of the variables in that equation. Choose the equation that is easier to work with. For example, if you have 3xy - y = 100 and x2 + y2 = 400, you should pick the first one since it can easily be solved for y very easily.

3xy - y = 100

y(3x - 1) = 100

y = 100/(3x - 1)

## Example 1, Step 2

Take the expression you obtained in Step 1 and plug it into the other equation. This will yield a single equation in one variable. For example, since we have y = 100/(3x - 1), we can transform the other equation as follows:

x2 + y2 = 400

x2 + [100/(3x - 1)]2 = 400

x2(3x - 1)2 + 10000 = 400(3x - 1)2

9x4 - 6x3 + x2 + 10000 = 3600x2 - 2400x + 400

9x4 - 6x3 - 3599x2 + 2400x + 9600 = 0

This is a quartic equation in x. You can use a graphing calculator or a quartic equation solving calculator to find the solutions for x. In this example, the solutions are

• x = 19.927518
• x = 2.008468
• x = -1.33707
• x = -19.932249

## Example 1, Step 3

Take the value(s) of x you found in Step 2 and plug them back into the expression from Step 1. This will give you the corresponding y values.

• x = 19.927518, y = 100/(3*19.927518 - 1) = 1.701185
• x = 2.008468, y = 100/(3*2.008468 - 1) = 19.898898
• x = -1.33707, y = 100/(3*(-1.33707) - 1) = -19.95526
• x = -19.932249, y = 100/(3*(-19.932249) - 1) = -1.644825

Thus, the four solution pairs are

• (19.927518, 1.701185)
• (2.008468, 19.898898 )
• (-1.33707, -19.95526 )
• (-19.932249, -1.644825 )

## Note

A system of non-linear equations in several variables can have 0, 1, 2, or more solutions depending on the form of the equations.

## Another Example

Let's use the substitution technique to solve the system of equations e^(xy) = 14035.5 and x^2 + y = 197.3, where x and y are restricted to the positive real numbers.

Solving the second equation for y gives you y = 197.3 - x^2.

Plugging this into the first equation gives you e^(197.3x - x^3) = 14035.5. This can be solved as follows:

e^(197.3x - x^3) = 14035.5
197.3x - x^3 = Ln(14035.5)
x^3 - 197.3x + 9.54934511 = 0

This cubic polynomial can be solved using techniques for cubic equations or with a numerical polynomial solver. The solutions to this equation are

x = -14.0705
x = 0.0484007
x = 14.0221

The corresponding y solutions for each x solution are

y = -0.67897
y = 197.29766
y = 0.68071

Discarding the first (x, y) pair because it is negative, we get the two solutions (0.0484007, 197.29766), and (14.0221, 0.68071).

## More by this Author

calculus-geometry 3 years ago from Germany

Nice example and explanation of how to solve non-linear systems. With polynomial equations you'll always get a finite number of solutions, but if you have functions like sine or cosine you can get infinitely many solutions.

paxwill 3 years ago from France Author

Thanks. I didn't want to use a non-linear example with trig functions for this reason.