**Solving Systems of Non-Linear Equations **

A “**system of equations**” is a collection of two or more equations that are solved simultaneously. Previously, I have gone over a few examples how to solve a system of **linear equations** using substitution and elimination methods. It is considered linear system because all the equations in the set are lines.

On the other hand, a **non-linear system** is a collection of equations that may contain some equations of line, but not all of them. In this lesson, we will only deal with system of non-linear equations with two equations in two unknowns, *x* and *y*.

There are seven (7) examples in this lesson.

**Example 1:** Solve the system of nonlinear equations below.

This system has two equations of each kind: a linear and a nonlinear. Start with the first equation since it is linear. You can solve for *x* or *y*. For this one, let’s solve for *y* in terms of *x*.

Substitute the value of *y* into the second equation, and then solve for *x*. In this problem, move everything to one side of the equation while keeping the opposite side equal to zero. After doing so, factor out the simple trinomial, and then set each factor equal to zero to solve for *x*.

After solving the equation, we arrived at two values of *x*. Substitute these numerical values to any of the two original equations. However, pick the “simpler” equation to simplify the calculation. Obviously, the linear equation ** x + y = 1** is the best choice!

- If
, solve for*x*= −3*y*

Answer:** (–3, 4)**

- If
, solve for*x*= 2*y*

Answer: **(2, –1)**

Therefore, the solution to the given system of nonlinear equations consist of two points which are **(–3, 4)** and **(2, –1)**.

Graphically, we can think of the solution to the system as the points of intersections between the linear function (*x* + *y* = 1) and quadratic function (*y* = *x*^{2 }− 5).

**Example 2:** Solve the system of equations below.

The first equation is a circle with a radius of 3 since the general formula of a circle is ** x^{2} + y^{2} = r^{2}**.

What I will do is to substitute the value of “*y*” of the bottom equation to the “*y*” of the top equation. Then we should be able to solve for *x*.

Use these values of *x* to find *y*. I would pick the bottom equation *y* = *x* + 3 to solve for *y*.

- If
, solve for*x*= 0*y*

Answer: **(0, 3)**

- If
, solve for*x*= −3*y*

Answer: **(–3, 0)**

The final answers are the points **(0, 3)** and **(–3, 0)**. These are the points of intersections of the given line and circle centered at the origin.

**Example 3:** Solve the system of equations below.

This problem is very similar to problem #2. We have a line (top equation) that intersects a circle (bottom equation) at two points.

**Step 1**: Solve the top equation for *y*.

**Step 2**: Plug in the value of *y* into the bottom equation. You will be required to square a binomial, combine like terms and factor out a trinomial to get the values of *x*. Here is the solution:

Therefore, the values of *x* are

**Step 3**: Back substitute these *x*-values into the top equation** x + y = –1** to get the corresponding

*y*-values.

Answer: **(–3, 2)**

Answer: **(2, –3)**

**Step 4**: Here is the graph of the line intersecting the circle at **(–3, 2)** and **(2, –3)**.

**Example 4:** Solve the system of nonlinear equations

Substitute the value of “*y* =” of the top equation to the bottom equation. Apply the distributive property then move everything to the left. Factor out the trinomial then set each factor equal to zero to solve for *x*.

So then we have…

Since we now have the values of *x*, pick any of the original equations to solve for *y*.

Answer: **(–1, 2)**

Answer: **(–2, 1)**

The graph shows the intersection of an oblique hyperbola and the line at points **(–1, 2)** and **(–2, 1)**.

**Example 5:** Solve the system of nonlinear equations

Observe that the first equation is of a circle centered at (-2, 2) with a radius of 1. The second equation is a parabola in standard form with vertex at (-2, 3). We expect that the solution to this system of nonlinear equations are the points of intersections of the given circle and parabola.

We will solve this two ways. First by substitution method then followed by elimination method.

**Example 6:** Solve the following system

Since the ** y^{2}** terms have the same coefficients but opposite in signs, we can add the two equations together to eliminate the variable y. This should leave us with a simple quadratic equation that can be solved easily using the square root method.

Next, divide both sides of the equation by the coefficient of the ** x^{2}** term , and followed by applying square root on both sides to get the values of x. Don’t forget to attach the plus or minus symbol whenever you get the square root of something.

Pick any of the two original equations, and find the values of y when ** x = ± 3**. I will use the first equation because it is much simpler!

- If
, solve for*x*= 3*y*

Answer: **(3, 1)** and **(3, –1)**

- If
, solve for*x*= –3*y*

Answer: **(–3, 1)** and **(–3, –1)**

The solution to this system of nonlinear equations consists of four points of intersections:

**(3, 1), (3, –1), (–3, 1) and (–3, –1)**

In fact, these are the points of intersections of the given ellipse (first equation) and hyperbola (second equation).

Graphically, it looks like this…

**Example 7:** Solve the following system

We will also solve this using elimination method. However, multiply first both of the equations by some number so that their constants become the same but opposite in signs.

Eliminate ** y^{2}** by multiplying the first equation by 2, and the second equation by 3 and finally adding them together!

Now, solve for *x* by dividing both sides by the coefficient of the ** x^{2}** term, and then performing square root operation in both sides of the equation.

Back substitute the values of *x* into any of the original equations to solve for *y*. Let’s use the first equation.

- If
, solve for*x*= 3*y*

Answer: **(3, 2)** and **(3, –2)**

- If
, solve for*x*= –3*y*

Answer: **(–3, 2)** and **(–3, –2)**

The solution to this nonlinear system are the points of intersections of the given ellipse and hyperbola.