Lesson Objectives
• Demonstrate an understanding of the substitution and elimination methods
• Learn how to solve a nonlinear system using substitution
• Learn how to solve a nonlinear system using elimination
• Learn how to solve a nonlinear system using a combination of substitution and elimination
• Learn how to solve a nonlinear system with an absolute value equation
• Learn how to solve a nonlinear system with non-real complex solutions

## How to Solve a System of Nonlinear Equations

A system of nonlinear equations is a collection of two or more equations, containing two or more variables, where at least one equation is nonlinear. To keep the math manageable, we will only focus on two-variable systems that contain only two equations. The strategy for solving these systems depends on the specific situation. In some cases, substitution works best, while in others, we would use elimination. We will also look at systems where a combination of substitution and elimination is used to obtain solutions. As a quick word of caution, when you are in this section, always check all solutions in the original equations of the system. Depending on the path you take, you may end up with extraneous solutions or solutions that do not satisfy the original equations.

### Solving a NonLinear System by Substitution

When one of the equations is linear, we begin by solving the linear equation for one of the variables. We can then use substitution to obtain our solutions. Let's look at an example.
Example #1: Solve each system. $$1) \, x^2 - y = 9$$ $$2) \, x + y = 3$$ We can see that equation #1 is a parabola and equation #2 is a line.
Solve equation #2 for y: $$x + y = 3$$ $$y = 3 - x$$ Plug in for y in equation #1: $$x^2 - y = 9$$ $$x^2 - (3 - x) = 9$$ $$x^2 - 3 + x = 9$$ Write the quadratic equation in standard form: $$x^2 + x - 12 = 0$$ Factor: $$(x + 4)(x - 3) = 0$$ $$x = -4, 3$$ Plug in for x in equation #2: $$x + y = 3$$ $$\text{if } x = -4$$ $$-4 + y = 3$$ $$y = 7$$ $$(-4, 7)$$ $$\text{if } x = 3$$ $$3 + y = 3$$ $$y = 0$$ $$(3, 0)$$ As we mentioned above, we should always check our solutions in this section. We already know both solutions will work in equation #2, so we just need to check both solutions in equation #1.
Check (-4, 7): $$x^2 - y = 9$$ $$(-4)^2 - 7 = 9$$ $$16 - 7 = 9$$ $$\require{color}9 = 9 \, \textcolor{green}{✓}$$ Check (3, 0): $$x^2 - y = 9$$ $$(3)^2 - 0 = 9$$ $$9 = 9 \, \textcolor{green}{✓}$$ For this problem, we can also verify our solutions graphically. Desmos Link for More Detail
$$x^2 - y = 9$$
$$x + y = 3$$

### Solving a NonLinear System by Elimination

For systems of linear equations, the elimination method is effective when both equations are in standard form (ax + by = c). For nonlinear systems, the elimination method is also effective when each equation is of the form ax2 + by2 = c. Let's look at an example.
Example #2: Solve each system. $$1) \, x^2 + y^2 = 52$$ $$2) \, x^2 + 4y^2 = 100$$ Multiply equation #1 by -1: $$-x^2 - y^2 = -52$$ $$x^2 + 4y^2 = 100$$ Add the equations: $$\begin{array}{rcl} -x^2 - y^2 = -52 \\ x^2 + 4y^2 = 100 \\ \hline 3y^2 = 48 \end{array}$$ Divide both sides by 3: $$y^2 = 16$$ Solve for y: $$y = \pm 4$$ Plug in for y in equation #1 (note: using 4 or -4 will give the same result here): $$x^2 + (4)^2 = 52$$ Solve for x: $$x^2 + 16 = 52$$ $$x^2 = 36$$ $$x = \pm 6$$ This leads to 4 solutions: $$(6, 4), (6, -4), (-6, 4), (-6, -4)$$ Since squaring a number or its opposite yields the same result, we only need to check one pair of values for x and y in each equation. In this case, we can check (6, 4) to verify all four solutions. $$x^2 + y^2 = 52$$ $$(6)^2 + (4)^2 = 52$$ $$36 + 16 = 52$$ $$52 = 52 \, \textcolor{green}{✓}$$ $$x^2 + 4y^2 = 100$$ $$(6)^2 + 4(4)^2 = 100$$ $$36 + 4(16) = 100$$ $$36 + 64 = 100$$ $$100 = 100 \, \textcolor{green}{✓}$$ For this problem, we can also verify our solutions graphically. Desmos Link for More Detail
$$x^2 + y^2 = 52$$
$$x^2 + 4y^2 = 100$$

### Solving a NonLinear System using a Combination of Substitution and Elimination

In some textbooks, you will see a type of problem that involves using both elimination and substitution. Let's look at an example.
Example #3: Solve each system. $$1) \, x^2 - xy + y^2 = 7$$ $$2) \, x^2 + 4xy + y^2 = 37$$ Multiply equation #1 by -1 to get equation #3: $$3) \, {-}x^2 + xy - y^2 = -7$$ Add equation #3 to equation #2 to get equation #4: $$\begin{array}{rcl} {-}x^2 + xy - y^2 = -7 \\ x^2 + 4xy + y^2 = 37 \\ \hline 5xy = 30 \end{array}$$ $$4) \, 5xy = 30$$ Solve equation #4 for y: $$y = \frac{30}{5x}, x ≠ 0$$ $$y = \frac{6}{x}, x ≠ 0$$ Plug in for y in equation #1: $$x^2 - xy + y^2 = 7$$ $$x^2 - x\left(\frac{6}{x}\right) + \left(\frac{6}{x}\right)^2 = 7$$ $$x^2 - 6 + \frac{36}{x^2} = 7$$ Multiply both sides by x2: $$x^4 - 6x^2 + 36 = 7x^2$$ $$x^4 - 13x^2 + 36 = 0$$ Factor: $$(x^2 - 9)(x^2 - 4) = 0$$ $$(x + 3)(x - 3)(x + 2)(x - 2) = 0$$ Solve for x: $$x = \pm 3, \pm 2$$ Plug in for x in equation #4: $$y = \frac{6}{x}$$ $$y = \frac{6}{-3} = -2$$ $$y = \frac{6}{3} = 2$$ $$y = \frac{6}{-2} = -3$$ $$y = \frac{6}{2} = 3$$ This leads to 4 solutions: $$(-3, -2), (3, 2), (-2, -3), (2, 3)$$ You really only have to check (3, 2) and (2, 3) if you consider the sign rules involved with the problem.
Check (3, 2): $$x^2 - xy + y^2 = 7$$ $$(3)^2 - (3)(2) + (2)^2 = 7$$ $$9 - 6 + 4 = 7$$ $$7 = 7 \textcolor{green}{✓}$$ $$x^2 + 4xy + y^2 = 37$$ $$(3)^2 + 4(3)(2) + (2)^2 = 37$$ $$9 + 24 + 4 = 37$$ $$37 = 37 \textcolor{green}{✓}$$ Check (2, 3): $$x^2 - xy + y^2 = 7$$ $$(2)^2 - (2)(3) + (3)^2 = 7$$ $$4 - 6 + 9 = 7$$ $$7 = 7 \textcolor{green}{✓}$$ $$x^2 + 4xy + y^2 = 37$$ $$(2)^2 + 4(2)(3) + (3)^2 = 37$$ $$4 + 24 + 9 = 37$$ $$37 = 37 \textcolor{green}{✓}$$ Desmos Link for More Detail
$$x^2 - xy + y^2 = 7$$
$$x^2 + 4xy + y^2 = 37$$

### Solving a NonLinear System with an Absolute Value Equation

Another common type of problem involves the absolute value operation. Let's look at an example.
Example #4: Solve each system. $$1) \, x^2 + y^2 = 36$$ $$2) \, |x| + y = 6$$ We will use the definition of the absolute value operation to split the problem up into two different cases. $$|x| = \begin{cases} x & \text{if } x ≥ 0 \\ -x & \text{if } x < 0 \end{cases}$$ Case #1: x is non-negative, this means |x| = x. $$3) \, x + y = 6$$ Case #2: x is negative, this means |x| = -x. $$4) \, {-}x + y = 6$$ Let's start with case #1, this means we will work with equations #3 and #1. $$1) \, x^2 + y^2 = 36$$ $$3) \, x + y = 6$$ Solve equation #3 for x: $$x = 6 - y$$ Plug in for x in equation #1: $$(6 - y)^2 + y^2 = 36$$ $$36 - 12y + y^2 + y^2 = 36$$ Solve for y: $$2y^2 - 12y = 0$$ $$2y(y - 6) = 0$$ $$2y = 0 \, \text{or} \, y - 6 = 0$$ $$2y = 0$$ $$y = 0$$ $$y - 6 = 0$$ $$y = 6$$ Plug in for y in equation #3: $$x + y = 6$$ $$\text{if } y = 0$$ $$x + 0 = 6$$ $$x = 6$$ $$(6, 0)$$ $$\text{if } y = 6$$ $$x + 6 = 6$$ $$x = 0$$ $$(0, 6)$$ For case #1, we have the following solutions: $$(6, 0), (0, 6)$$ For case #2, we will work with equations #1 and #4. $$1) \, x^2 + y^2 = 36$$ $$4) \, {-}x + y = 6$$ Solve equation #4 for x: $$-x = 6 - y$$ $$x = y - 6$$ Plug in for x in equation #1: $$(y - 6)^2 + y^2 = 36$$ $$y^2 - 12y + 36 + y^2 = 36$$ Solve for y: $$2y^2 - 12y = 0$$ We previously solved the above equation in case #1. $$y = 0, 6$$ Plug in for y in equation #4. $$-x + y = 6$$ $$\text{if } y = 0$$ $$-x + 0 = 6$$ $$x = -6$$ $$(-6, 0)$$ $$\text{if } y = 6$$ $$-x + 6 = 6$$ $$x = 0$$ $$(0, 6)$$ For case #2, we have the following solutions: $$(-6, 0), (0, 6)$$ Since (0, 6) is a duplicate, we have 3 distinct solutions for the system. $$(-6, 0), (0, 6), (6, 0)$$ Check: $$1) \, x^2 + y^2 = 36$$ $$2) \, |x| + y = 6$$ For (-6, 0) and (6, 0), the results will be the same since x2 and |x| are both non-negative. $$(6)^2 + 0^2 = 36$$ $$36 = 36 \, \textcolor{green}{✓}$$ $$|6| + 0 = 6$$ $$6 = 6 \, \textcolor{green}{✓}$$ $$(0)^2 + 6^2 = 36$$ $$36 = 36 \, \textcolor{green}{✓}$$ $$|0| + 6 = 6$$ $$6 = 6 \, \textcolor{green}{✓}$$ Desmos Link for More Detail
$$x^2 + y^2 = 36$$
$$|x| + y = 6$$

### Solving a NonLinear System with Non-Real Complex Solutions

In some scenarios, we may end up with non-real complex solutions. Let's look at an example.
Example #5: Solve each system. $$1) \, x^2 + y^2 = 8$$ $$2) \, 4x^2 + 3y^2 = 20$$ Multiply equation #1 by -3: $$3) \, {-}3x^2 - 3y^2 = -24$$ Add equations #2 and #3: $$\begin{array}{rcl} 4x^2 + 3y^2 = 20 \\ {-}3x^2 - 3y^2 = -24 \\ \hline x^2 = -4 \end{array}$$ Solve for x: $$x^2 = -4$$ $$x = \pm \sqrt{-4} = \pm 2i$$ Plug in for x in equation #1: $$\text{if } x = 2i$$ $$x^2 + y^2 = 8$$ $$(2i)^2 + y^2 = 8$$ $$(2)^2(i)^2 + y^2 = 8$$ $$i^2 = -1$$ $$-4 + y^2 = 8$$ $$y^2 = 12$$ Solve for y: $$y = \pm \sqrt{12} = \pm 2\sqrt{3}$$ $$\left(2i, 2\sqrt{3}\right), \left(2i, -2\sqrt{3}\right)$$ $$\text{if } x = -2i$$ $$x^2 + y^2 = 8$$ $$(-2i)^2 + y^2 = 8$$ $$(-2)^2(i)^2 + y^2 = 8$$ $$i^2 = -1$$ $$-4 + y^2 = 8$$ $$y^2 = 12$$ Solve for y: $$y = \pm \sqrt{12} = \pm 2\sqrt{3}$$ $$\left(-2i, 2\sqrt{3}\right), \left(-2i, -2\sqrt{3}\right)$$ The solutions for the system are: $$\left(2i, 2\sqrt{3}\right), \left(2i, -2\sqrt{3}\right), \left(-2i, 2\sqrt{3}\right), \left(-2i, -2\sqrt{3}\right)$$ Since we have both x2 and y2 in each equation, we only have to check one of our solutions.
Check: $$1) \, x^2 + y^2 = 8$$ $$2) \, 4x^2 + 3y^2 = 20$$ $$(2i)^2 + \left(2\sqrt{3}\right)^2 = 8$$ $$-4 + 12 = 8$$ $$8 = 8 \, \textcolor{green}{✓}$$ $$4(2i)^2 + 3\left(2\sqrt{3}\right)^2 = 20$$ $$-16 + 36 = 20$$ $$20 = 20 \, \textcolor{green}{✓}$$ Since all of our solutions here involve imaginary numbers, we will not be able to show the solution graphically. This is because imaginary numbers are not represented on the Cartesian Coordinate plane, which represents real number coordinates for x and y. Desmos Link for More Detail
$$x^2 + y^2 = 8$$
$$4x^2 + 3y^2 = 20$$

#### Skills Check:

Example #1

Solve each system of equations. $$x^2 + y^2 + 12x - 2y = -17$$ $$2x - y = -3$$

A
$$(2, 1)$$
B
$$(-3, -2)$$
C
$$(5, -4)$$
D
$$(-2, -1)$$
E
$$(-2, 1)$$

Example #2

Solve each system of equations. $$x^2 - 10x + y = -30$$ $$x^2 - 5x + y = 0$$

A
$$(-6, -6)$$
B
$$(2, -2)$$
C
$$(-2, -2)$$
D
$$(6, -6)$$
E
$$(0, -5)$$

Example #3

Solve each system. $$xy = -10$$ $$3x - 2y = -16$$

A
$$(5, 2), (3, -4)$$
B
$$\left(\frac{1}{2}, -2\right), (-2, 4)$$
C
$$(-2, 5), \left(-\frac{10}{3}, 3\right)$$
D
$$(5, -2), \left(-4, -\frac{1}{3}\right)$$
E
$$(-6, -3), (1, 0)$$