Solving Rational Equations-General Mathematics

A helpful presentation for the discussion of Rational Equation

1. Solving Rational
Equations
xx

4
11
x2

2. Rational Equation
An equation containing one or
more rational expressions
2
2 1 5
3
x x
x x
 
 

3. Steps to solve Rational Equations
1.Find the LCD
2.Multiply every term on both sides of
the equation by the LCD over 1
3.Cross Cancel
4.Solve for the variable
a. Linear equation: get variables on
one side and constants on the other
b. Quadratic: set your equation = 0
and factor.

4. Extraneous Solutions
•When both sides of the equation are
multiplied by a variable, the equation is
transformed into a new equation and
may have an extra solution.
•Check each solution in the original
rational equation
•Make sure that your answer does not
make the denominator 0

5. Solving Rational EquationsSolving Rational Equations
Multiply both sides of the equation
by the LCM of the denominators.
xx

4
11
Least Common Multiple: Each
factor raised to the greatest
exponent.
xx4
LCM is 4x x
xx4
xx4
x24 x2

6. Solve for x:
xx
2
1
1


LCM =
x
(x + 1)(x)
(x + 1)(x)• •(x + 1)(x)
22x
x2
1x 1x
0 2x

7. Solve for x:
12
1
2
1

xx
LCM =
21
2x
2x• •2x
x24
x
8
1

8. Solve for x:
2
1

x
x
1
2
x
x• • x
x2
1x
012
2
xx
01
2
x

9. Solve for x:
2
4
2
2


xx
x
2x
4
2
x
2x
-2 is an extraneous solution.

10. Solve for x:
128
5
4
3 x
 LCM =
x21518 
x
2
33
2
2
3
2
3
2 3 24 
24
1
24
1

2
2 3

11. Solve for x:
2
4
2
2


xx
x
2x
4
2
x
2x
-2 is an extraneous solution.

12. Cross products
Short cut:
7 1
2 4
x
x



4 7x 
extraneous solution?
1 2x
4 28 2x x 
1x 1x
3 28 2x 
3 30x
10x

13. Cross products:2 1
1 2x x

 
1 2 4x x 
5x

14. Cross products:3 1
3 2 5
x
x



5 15 3 2x x  
2 17x
17
2
x

15. Cross products:2 1
3 1
x x
x x
 

 
2
2x x  
2
4 3x x 
2
x
2
x
2 4 3x x   
3 2 3x  5
3
x
4x 4x
3 5x

16. Assignment:
Page 453
# (2-24) even