Unformatted text preview: ay 6: RATIONAL EQUATIONS (Part 1) RATIONAL EQUATIONS
A rational equation has at least one term that is a rational expression.
Rational equations can be simplified in two ways.
1. Rewrite each side of the equation as a single fraction with common denominators.
Compare each fraction; the numerators must be equal.
2. Multiply each side of the equation by the lowest common denominator.
Roots must always checked against the restrictions from the rational expressions. example: Solve 1 − 3 = 1− 2
2x
8
x o Determine the restrictions.
o Multiply both sides by the LCD, 8x.
o Distribute. 1 − 3 = 1− 2
2x
8
x x≠ 0 3
( 8 ) = 8x ( 1 − 2 )
x
3
( ) ( 8 ) = 8x(1) − 8x ( 2 )
x 8x 1 −
2x
8 x 1 − 8x
2x 4 − 3x o Multiply. = 20
11 x = 20
11 o Check this root against the restriction x ≠ 0
Answer: x
=1
3x − 9
x−3 x
1
=
3( x − 3)
x−3
x≠ 3 o Determine the restrictions.
o Multiply by the LCD, 3(x − 3).
o Solve. 8x − 1 6 x o Continue solving. example: Solve = 3(x − 3) x = 3(x − 3) 1 3( x − 3) x − 3 x=3 o Check this root against the restriction x ≠ 3; this root is extraneous.
Answer: This equation has no solution. solution set = φ Unit 5: Day 6 notes  Rational Equations (Part 1) Page 2 of 2 x−6=
12
x − 3 x + 1 x2 − 2 x − 3 example: Solve x−6=
12
x−3
x +1
( x − 3)( x + 1) x ≠ −1 , x ≠ 3
12 (x + 1)(x − 3) x − 6 = (x + 1)(x − 3) ( x − 3)( x + 1) x − 3 x + 1 (x + 1)(x − 3) x
( x − 3) − (x + 1)(x − 3) ( x 6 1)
+ 12 = (x + 1)(x − 3) ( x − 3)( x + 1) x (x + 1) − 6(x − 3) = 12
x2 + x − 6x + 18 = 12
x2 − 5x + 18 = 12
x 2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
x−2 = 0 or x−3 = 0 x= 2 or x= 3 Check these roots against the restrictions x ≠ −1 and x ≠ 3 ; 3 is extraneous.
Answer: x=2 exercises: Solve.
a) 4 + 6 = 2
x c) x+2
3
−x=
2x + 1
3
4x + 2 y
3
=
y+3
2y − 5 b) 1− d) z +2
+ 23
=1
z+3
z
z + 3z 3
[Answers: −1, 8, 2 , −1] PRECALCULUS 11 Unit 5 – Day 7: RATIONAL EQUATIONS (Part 2) SOLVING PROBLEMS WITH RATIONAL EQUATIONS
Solving problems using algebra:
1) Define variables introduced and write expressions for all important quantities.
2) Write the system of equations that models the situation described in the problem.
3) Solve the system of equations.
4) Answer the problem. Your answer must be what the problem is asking you to find. example: One cyclist averages 3 km/h faster than a second cyclist. The faster cyclist
rode 270 km in the same time the slower cyclist rode 225 km. What was the
average speed of each cyclist? 1) s will represent the speed of the first cyclist (in km/h).
s − 3 then represents the speed of the second cyclist (in km/h).
270 is then the time the first cyclist rode (in hours)
s
225 is then the time the second cyclist rode (in hours)
s−3 2) 3) 4) The cyclists rode for the same amount of time, therefore s(s − 3) 270 s
270(s − 3)
270s − 810
4 5s
s
s−3 270 = 225
s
s−3
= s(s − 3) 225 s − 3
= 225s
= 225s
= 810
= 18
= 15 The first cyclist averaged 18 km/h and the second cyclist averaged 15 km/h. Unit 5: Day 7 notes  Rational Equations (Part 2) Page 2 of 2 exercise: The speed of the current in a river was 2 km/h. A boat made a round trip to a
town 24 km away in a total of 5 hours. What is the speed of the boat in still
water? [Answer: 10 km/h] exercise: Find the value(s) of x in the numbers x , x + 1 , and x + 2 such that the
reciprocal of the smallest number equals the sum of the reciprocals of the
other two numbers. [Answer: − 2 or 2...
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 Fall '11
 Aytona
 Calculus, PreCalculus, Rational Expressions, Fractions, Integers, ........., Elementary arithmetic

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