Solving Rational Equations

Have a polynomial in the numerator and a polynomial in the denominator

ex.

1. Factor the DENOMINATORS only

2. Multiply each side of the equation by the least common multiple (LCM) this should get rid of the fraction

3. Solve

4. Check to make sure your answers work (you don't divide by zero)

You don't need to factor the denominators.  

Factors: 3, x-2

Multiply each side by (3)(x-2)

 $\left(3\right)\left(x-2\right)\cdot\frac{x}{3}=\frac{1}{x-2}\cdot\left(3\right)\left(x-2\right)$  

 $\left(x-2\right)\cdot x=1\cdot\left(3\right)$  

 $x^2-2x=3$  

 $x^2-2x-3=0$  

(x - 3)(x + 1) = 0

x - 3 = 0 or x + 1 = 0

x = 3 or x = -1

 $\frac{x}{3}=\frac{1}{x-2}$ 

The only denominator with x is x - 2

 $x\ -\ 2\ne0$  

 $x\ne2$  

Your solutions are good

LCM:  (5r  + 6)(6)

 $\left(5r+6\right)\left(6\right)\cdot\frac{4r}{5r+6}=\frac{1}{6}\cdot\left(5r+6\right)\left(6\right)$  

(6)(4r) = (1)(5r+6)

24r=5r + 6

19r = 6

 $r\ =\ \frac{6}{19}$  

You need to make sure that you are NOT dividing by zero!

 $\frac{4r}{5r+6}=\frac{1}{6}\$  

 $5r+6\ne0$  

 $5r\ne-6$  

 $r\ne-\frac{6}{5}$  

your solution is good

LCM:  (x-3)(x+5)

 $\left(x-3\right)\left(x+5\right)\cdot-\frac{x^2+x-8}{\left(x-3\right)\left(x+5\right)}=\frac{x}{x-3}\cdot\left(x-3\right)\left(x+5\right)$  

 $-\left(x^2+x-8\right)=\left(x\right)\left(x+5\right)$  

 $-x^2-x+8=x^2+5x$  

 $0=2x^2+6x-8$  

 $0=x^2+3x-4$  

0 = (x + 4)(x - 1)

x + 4 = 0 or x - 1 = 0

x = -4 or x = 1

 $-\frac{x^2+x-8}{\left(x-3\right)\left(x+5\right)}=\frac{x}{x-3}$  

 $x\ -\ 3\ne0\ \ \ and\ x\ +\ 5\ne0$  

 $x\ne3\ and\ x\ne-5$  

Your solutions are good

LCM:  (x - 3)(x - 1)

 $\left(x-3\right)\left(x-1\right)\cdot\frac{x+1}{x-3}=-\frac{x^2-6x+1}{\left(x-3\right)\left(x-1\right)}\cdot\left(x-3\right)\left(x-1\right)$  

 $\left(x-1\right)\left(x+1\right)=-\left(x^2-6x+1\right)$  

 $x^2+x-x-1=-x^2+6x-1$  

 $x^2-1=-x^2+6x-1$  

 $2x^2-6x=0$  

2x(x - 3) = 0

2x = 0 or x - 3 = 0

x = 0 or x = 3

 $x-3\ne0\ and\ x\ -1\ \ne0$  

 $x\ne3\ and\ x\ne1$  

The only solution that works is x = 0

Remember to find an inverse you switch x and y and then solve for y?

Find the inverse  $\left(f^{-1}\left(x\right)\right)$  of  $f\left(x\right)=\frac{6x-5}{x+9}$  

Make f(x) = y:   $y=\frac{6x-5}{x+9}$  

Switch x and y:   $x=\frac{6y-5}{y+9}$  

Solve for y:  $\left(x\right)\left(y+9\right)=6y-5$  

xy + 9x = 6y - 5

Get all of the y's to one side of the equation: 9x = 6y - 5 - xy 

Isolate y:  9x + 5 = 6y - xy

Factor the y out on the right side:  9x + 5 = y(6 - x)

Solve for y:   $\frac{9x+5}{6-x}=y$  

Write with inverse notation:   $f^{-1}\left(x\right)=\frac{9x+5}{6-x}$  

 $y=\frac{5+x}{4-2x}$  

 $x=\frac{5+y}{4-2y}$  

 $\left(x\right)\left(4-2y\right)=5+y$  

4x - 2xy = 5 + y

4x = 5 + y + 2xy

4x - 5 = y + 2xy

4x - 5 = y(1 + 2x)

 $y=\frac{4x-5}{1+2x}$  

 $h^{-1}\left(x\right)=\frac{4x-5}{1+2x}$  

1. Rational Equations intro

2. Rational equations

3. Find inverses of rational functions

Due tomorrow 11:59 pm