Solving Rational Equations

Solving Rational Equations

Rational expressions

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

• ex.

   $\frac{x^2+3x+6}{2x^2+4x}$  

Solving rational equations

• 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)

Solve

 $\frac{x}{3}=\frac{1}{x-2}$  

• 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

Check your answers 

•  $\frac{x}{3}=\frac{1}{x-2}$ 

• The only denominator with x is x - 2

•  $x\ -\ 2\ne0$  
  

•  $x\ne2$  

• Your solutions are good

Fill in the Blank

Solve

 $\frac{4r}{5r+6}=\frac{1}{6}$  Write your answer as a reduced fraction.

 $\frac{4r}{5r+6}=\frac{1}{6}$  

• 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

•  $r\ =\ \frac{6}{19}$  

Check

• 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

Solve

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

• 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

Check:

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

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

•  $x\ne3\ and\ x\ne-5$  

• Your solutions are good

Multiple Select

Solve

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

• -5

• -2

• 0

• 1

• 4

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

• 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

check

•  $x-3\ne0\ and\ x\ -1\ \ne0$  

•  $x\ne3\ and\ x\ne1$  

• The only solution that works is x = 0

Inverse

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

Multiple Choice

Find

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

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

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

• none of these

 $h\left(x\right)=\frac{5+x}{4-2y}$  

•  $y=\frac{5+x}{4-2x}$  

•  $x=\frac{5+y}{4-2y}$  

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

• 4x - 2xy = 5 + y

• 4x = 5 + y + 2xy

• 4x - 5 = y + 2xy

• 4x - 5 = y(1 + 2x)

•  $y=\frac{4x-5}{1+2x}$  

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

Homework in Khan Academy

• 1. Rational Equations intro

• 2. Rational equations

• 3. Find inverses of rational functions

• Due tomorrow 11:59 pm