## Wednesday, June 8, 2016

### High School Math Solutions – Systems of Equations Calculator, Nonlinear

In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how to solve a system of nonlinear equations. A system of nonlinear equations is a system in which one or more variables has an order greater than 1, and/or there is a product of variables in one of the equations. There are two ways to solve systems of nonlinear equations, substitution and elimination.

### Substitution

Very similar to the method of substitution for systems of linear equations

1. Solve for one variable in one of the equations
2. Substitute the value of that variable in another equation
3. Solve for variable in that equation
4. Solve for other variables using known variable value
5. Double check to make sure the ordered pairs work by plugging them back into an equation

### Elimination

Also very similar to the method of substitution for systems of linear equations

1. Multiply by a constant
2. Add/subtract to get rid of a variable
3. Solve for the variable
4. Solve for other variables using known variable value
5. Double check to make sure the ordered pairs work by plugging them back into an equation

It is a lot easier to visualize these two methods than to just read about them. I will go through two examples step by step, one for each method.

3-x^2=y
x+1-y=0
1. Solve for y in the second equation
y=x+1
2. Substitute x+1 for y in the first equation
3-x^2=x+1
3. Solve for x
x^2+x-2=0
(x+2)(x-1)=0
x=1\:or\:x=-2
4. Solve for y by substituting in the values for x
y=x+1
y=(1)+1 \:\:\:\: y=(-2)+1
y=2 \:\:\:\:\:\: y=-1

5. Double check
(1,2) \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: (-2,-1)
3-(1)^2=2 \:\:\:\:\:\:\:\:\:\:\: 3-(-2)^2=-1
2=2 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-1=-1

x^2+y=5
x^2+y^2=7
1. We don’t need to multiply by a constant because x^2 already has the same constant of 1 in both equations.

2. Subtract to get rid of x^2
\:\:\:x^2+y=5
-\:x^2+y^2=7
-------
y-y^2=-2
3. Solve for y
y^2-y-2=0
(y-2)(y+1)=0
y=2\:or\:y=-1
4. Solve for x
y=2 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: y=-1
x^2+(2)=5 \:\:\:\:\:\:\:\:\:\: x^2+(-1)=5
x^2=3 \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: x^2=6
x=\pm\sqrt{3} \:\:\:\:\:\:\:\:\:\:\:\:\:\: x=\pm\sqrt{6}
5. Double check