Download Solving Rational Equations

Solving Rational Equations
To solve an equation that has one or more fractions, we found the LCD
and cleared the fraction(s) by multiplying each term by the LCD. This is
the same strategy we will use to solve equations that have one or more
rational equations.
Recall that we cannot divide by zero, so a fraction with a zero
denominator is referred to as undefined. With equations involving
rational equations, we will need to check that the final answer will not
make the denominator zero. If it does, that answer is not a solution
which is called an extraneous solution.
Example 1: Solve:
1
2x
7
4
6
x
+ =
The LCD: 6x
(6x)
1
2x
7
4
6
x
+ (6x) =
(6x)
3 + 7x = 24
-3
-3
7x = 21
7
7
x=3
Multiply each term by the LCD
Simplified equation
Subtract 3
Divide by 7
3 will not make any of the denominators zero, so it is valid
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 2: Solve:
5x+5
x+2
+ 3x =
x2
x+2
The LCD: (x + 2)
(x + 2)
5x+5
+ 3x(x + 2) =
x+2
x2
x+2
(x + 2) Multiply each term by the LCD
5x + 5 + 3x2 + 6x = x2
3x2 + 11x + 5 = x2
- x2
- x2
2x2 + 11x + 5 = 0
(2x + 1)(x + 5) = 0
2x + 1 = 0 and x + 5 = 0
-1–1
-5–5
2x
= -1
x = -5
2
2
x=-
Simplified equation
Combine like terms
Subtract x2
Factor
Set each factor equal to zero
Solve resulting equations
1
2
1
3
- +2= and -5+2= -3
2
2
neither solution made a denominator 0, so both are valid
Example 3: Solve:
1
x−3
1
x−3
+
+
2x
x2 −9
=
1
x+3
2x
(x−3)(x+3)
=
1
x+3
Factor the denominator
The LCD: (x – 3)(x + 3)
(x – 3)(x + 3)
1
x−3
+
2x
(x−3)(x+3)
(x – 3)(x + 3) =
1
x+3
(x – 3)(x + 3)
Multiply each term by the LCD
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
x + 3 + 2x = x – 3
3x + 3 = x – 3
-x
-x
2x + 3 = - 3
-3
-3
2x
= -6
2
2
x = -3
Simplified equation
Combine like terms
Subtract x
Subtract 3
Divide by 2
-3-3= -6 and -3+3= 0
-3 made one of the denominators 0, so extraneous solution
x
Example 4: Solve:
–
x−1
x
x−1
–
1
x−2
1
x−2
=
=
11
x2 −3x+2
11
(x−1)(x−2)
Factor the denominator
The LCD: (x – 1)(x – 2)
(x – 1)(x – 2)
x
x−1
–
1
x−2
(x – 1)(x – 2) =
11
(x−1)(x−2)
(x – 1)(x – 2)
Multiply each term by the LCD
x(x – 2) – 1(x – 1) = 11
x2 – 2x – x + 1 = 11
x2 – 3x + 1 = 11
- 11 – 11
2
x – 3x – 10 = 0
(x – 5)(x + 2) = 0
x – 5 = 0 and x + 2 = 0
+5+5
-2–2
x=5
x = -2
Simplified equation
Distribute
Combine like terms
Subtract 11
Factor the polynomial
Set each factor equal to zero
Solve resulting equations
5-1= 4 and 5-2= 3 and -2-1= -3 and -2-2= -4
neither solution made a denominator 0, so both are valid
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
x
Example 5: Solve:
x−1
–
2
x
=
1
x−1
The LCD: x(x – 1)
x(x – 1)
x
x−1
–
2
x
x(x – 1) =
1
x−1
x2 – 2(x – 1) = 1x
x2 – 2x + 2 = x
-x
-x
2
x – 3x + 2 = 0
(x – 2)(x – 1) = 0
x – 2 = 0 and x – 1 = 0
+2+2
+1+1
x=2
x=1
x(x – 1) Multiply each term by the LCD
Simplified equation
Distribute
Subtract x from both sides
Factor the polynomial
Set each factor equal to zero
Solve resulting equations
2-1=1 and 1-1=0
2 does not make a denominator 0, so it is a solution
1 does make a denominator 0, so it an extraneous solution
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)