Lesson Objectives

- Demonstrate an understanding of the substitution and elimination methods
- Learn how to solve a non-linear system using substitution
- Learn how to solve a non-linear system using elimination
- Learn how to solve a non-linear system using a combination of substitution and elimination

## How to Solve a System of Non-Linear Equations

We previously learned how to solve a system of linear equations using graphing, substitution, and elimination. In this lesson, we will learn how to
solve non-linear systems. An equation where some terms have two or more variables multiplied together or a variable that contains an exponent of 2 or
larger is referred to as a non-linear equation. A system of non-linear equations includes two or more equations where at least one of the equations is
non-linear.

Example 1: Solve each non-linear system $$x^2 + y^2 = 4$$ $$x + y = -2$$ Let's solve our second equation for y: $$y = -x - 2$$ Now, we can plug in for y in the first equation: $$x^2 + (-x - 2)^2 = 4$$ $$x^2 + x^2 + 4x + 4 = 4$$ $$2x^2 + 4x = 0$$ Solve by factoring: $$2x(x + 2) = 0$$ $$2x = 0$$ $$x = 0$$ $$x + 2 = 0$$ $$x = -2$$ This tells us x can be 0 or x can be -2. We need to plug each one back in for x in one of the original equations. We can use either one. Since it is easier to work with the linear equation, let's use equation 2: $$0 + y = -2$$ $$y = -2$$ This gives us the ordered pair (0,-2) $$-2 + y = -2$$ $$y = 0$$ This gives us the ordered pair (-2,0)

Our two solutions are:

x = 0, y = -2 (0,-2)

x = -2, y = 0 (-2,0)

Example 2: Solve each non-linear system $$3x^2 + 2y^2 = 12$$ $$x^2 + 2y^2 = 4$$ We can multiply the second equation by -1. This will give us opposite coefficients on y

Our two solutions are:

x = 2, y = 0 (2,0)

x = -2, y = 0 (-2,0)

### Solving a Non-Linear System by Substitution

Let's look at an example that can be solved using the substitution method.Example 1: Solve each non-linear system $$x^2 + y^2 = 4$$ $$x + y = -2$$ Let's solve our second equation for y: $$y = -x - 2$$ Now, we can plug in for y in the first equation: $$x^2 + (-x - 2)^2 = 4$$ $$x^2 + x^2 + 4x + 4 = 4$$ $$2x^2 + 4x = 0$$ Solve by factoring: $$2x(x + 2) = 0$$ $$2x = 0$$ $$x = 0$$ $$x + 2 = 0$$ $$x = -2$$ This tells us x can be 0 or x can be -2. We need to plug each one back in for x in one of the original equations. We can use either one. Since it is easier to work with the linear equation, let's use equation 2: $$0 + y = -2$$ $$y = -2$$ This gives us the ordered pair (0,-2) $$-2 + y = -2$$ $$y = 0$$ This gives us the ordered pair (-2,0)

Our two solutions are:

x = 0, y = -2 (0,-2)

x = -2, y = 0 (-2,0)

### Solving a Non-Linear System by Elimination

When two-second degree equations are involved, we normally use the elimination method. Let's take a look at an example.Example 2: Solve each non-linear system $$3x^2 + 2y^2 = 12$$ $$x^2 + 2y^2 = 4$$ We can multiply the second equation by -1. This will give us opposite coefficients on y

^{2}: $$3x^2 + 2y^2 = 12$$ $$-x^2 - 2y^2 = -4$$ Now, we can add the left sides together and set this equal to the sum of the right sides. $$\require{cancel}3x^2 + \cancel{2y^2} = 12$$ $$\underline{-x^2 + \cancel{-2y^2} = -4}$$ $$2x^2 = 8$$ Divide each side by 2: $$x^2 = 4$$ Solve using the square root property: $$x = \pm \sqrt{4}$$ $$x = 2,-2$$ Now, we can plug these solutions back in for x in one of the original equations: $$3(2)^2 + 2y^2 = 12$$ $$12 + 2y^2 = 12$$ $$2y^2 = 0$$ $$y^2 = 0$$ $$y = 0$$ This gives us the ordered pair (2,0) $$3(-2)^2 + 2y^2 = 12$$ $$12 + 2y^2 = 12$$ $$2y^2 = 0$$ $$y^2 = 0$$ $$y = 0$$ This gives us the ordered pair (-2,0)Our two solutions are:

x = 2, y = 0 (2,0)

x = -2, y = 0 (-2,0)

### Solving a Non-Linear System using a Combination of Substitution and Elimination

$$-2y^2 + 3xy + 2 = 0$$ $$4y^2 + 9x^2 - 10 = 0$$ To solve this equation, we will use a combination of the elimination and substitution methods. Let's multiply the first equation by 2. This will give us opposite coefficients (-4 and 4) on y^{2}. $$-4y^2 + 6xy + 4 = 0$$ $$4y^2 + 9x^2 - 10 = 0$$ Let's move our constants to the right side: $$-4y^2 + 6xy = -4$$ $$4y^2 + 9x^2 = 10$$ Now, we will add the left sides and set this equal to the sum of the right sides: $$\cancel{-4y^2} + 6xy = -4$$ $$\underline{\cancel{4y^2} + 9x^2 = 10}$$ $$9x^2 + 6xy = 6$$ We can divide each side by 3: $$3x^2 + 2xy = 2$$ Now, we will solve the equation for y: $$2xy = -3x^2 + 2$$ $$y = \frac{-3x^2 + 2}{2x}$$ Now, we will substitute in for y in one of the original equations. Using the second equation will be a bit easier. $$4y^2 + 9x^2 - 10 = 0$$ $$4\left(\frac{-3x^2 + 2}{2x}\right)^2 + 9x^2 - 10 = 0$$ $$4\frac{(-3x^2 + 2)^2}{4x^2} + 9x^2 - 10 = 0$$ $$\cancel{4}\frac{(9x^4-12x^2+4)}{\cancel{4}x^2} + 9x^2 - 10 = 0$$ $$\frac{(9x^4-12x^2+4)}{x^2} + 9x^2 - 10 = 0$$ We can multiply each side by x^{2}to clear the denominator: $$9x^4-12x^2+4 + 9x^4 - 10x^2 = 0$$ $$18x^4-22x^2+4= 0$$ We can divide each side of the equation by 2: $$9x^4 - 11x^2 + 2 = 0$$ This equation is quadratic in form and can be solved by factoring: $$(x + 1)(x - 1)(9x^2 - 2) = 0$$ $$x + 1 = 0$$ $$x = -1$$ $$x - 1 = 0$$ $$x = 1$$ $$9x^2 - 2 = 0$$ $$9x^2 = 2$$ $$x^2 = \frac{2}{9}$$ $$x = \pm \sqrt{\frac{2}{9}}$$ $$x = \pm \frac{\sqrt{2}}{3}$$ So we have four values for x: $$x = \pm 1, \pm\frac{\sqrt{2}}{3}$$ To find y, we will use our equation: $$y = \frac{-3x^2 + 2}{2x}$$ If x is 1: $$y = \frac{-3(1)^2 + 2}{2(1)}$$ $$y = \frac{-3 + 2}{2}$$ $$y = \frac{-1}{2}$$ $$x = 1, y = \frac{-1}{2}$$ $$\left(1,\frac{-1}{2}\right)$$ If x is -1: $$y = \frac{-3(-1)^2 + 2}{2(-1)}$$ $$y = \frac{-3 + 2}{-2}$$ $$y = \frac{-1}{-2}$$ $$x = -1, y = \frac{1}{2}$$ $$\left(-1,\frac{1}{2}\right)$$ If x is square root of 2 over 3: $$y = \frac{-3\left(\frac{\sqrt{2}}{3}\right)^2 + 2}{2\left(\frac{\sqrt{2}}{3}\right)}$$ $$y = \frac{-3\left(\frac{2}{9}\right) + 2}{\frac{2\sqrt{2}}{3}}$$ $$y = \frac{\frac{-2}{3} + 2}{\frac{2\sqrt{2}}{3}}$$ $$y = \frac{\frac{4}{3}}{\frac{2\sqrt{2}}{3}}$$ $$y = \frac{2\cancel{4}}{\cancel{3}} \cdot \frac{\cancel{3}}{\cancel{2}\sqrt{2}}$$ $$y = \frac{2}{\sqrt{2}}$$ If we rationalize the denominator we end up with: $$y = \sqrt{2}$$ $$x = \frac{\sqrt{2}}{3}, y = \sqrt{2}$$ $$\left(\frac{\sqrt{2}}{3},\sqrt{2}\right)$$ If x is the negative square root of 2 over 3: $$y = \frac{-3\left(\frac{-\sqrt{2}}{3}\right)^2 + 2}{2\left(\frac{-\sqrt{2}}{3}\right)}$$ $$y = \frac{-3\left(\frac{2}{9}\right) + 2}{\frac{-2\sqrt{2}}{3}}$$ $$y = \frac{\frac{-2}{3} + 2}{\frac{-2\sqrt{2}}{3}}$$ $$y = \frac{\frac{4}{3}}{\frac{-2\sqrt{2}}{3}}$$ $$y = \frac{2\cancel{4}}{\cancel{3}} \cdot \frac{\cancel{3}}{-1\cancel{2}\sqrt{2}}$$ $$y = \frac{2}{-\sqrt{2}}$$ If we rationalize the denominator we end up with: $$y = -\sqrt{2}$$ $$x = -\frac{\sqrt{2}}{3}, y = -\sqrt{2}$$ $$\left(-\frac{\sqrt{2}}{3},-\sqrt{2}\right)$$
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