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Please complete the Prerequisite Skills
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

Big ideas:

Performing operations with rational expressions

Solving rational equations




What are the differences
between direct, inverse and joint
variation?


 Inverse
variation ~ The relationship of two
variables x and y if there is a nonzero number
𝑎
a such that y =
𝑥
 Constant
of variation ~ The nonzero constant a
in a direct variation equation y=ax, an inverse
𝑎
variation equation y= , or a joint variation
𝑥
equation z=axy
 Joint variation ~ A relationship that occurs
when a quantity varies directly with the
product of two or more other quantities.
EXAMPLE 1
Classify direct and inverse variation
Tell whether x and y show direct variation, inverse
variation, or neither.
Given Equation
a. xy = 7
Rewritten Equation
y=
b. y = x + 3
c.
y
= x
4
7
x
Type of Variation
Inverse
Neither
y = 4x
Direct
EXAMPLE 2
Write an inverse variation equation
The variables x and y vary inversely, and y = 7 when
x = 4. Write an equation that relates x and y. Then find y when x = –2 .
y=
7=
a
x
a
4
28 = a
Write general equation for inverse variation.
Substitute 7 for y and 4 for x.
Solve for a.
ANSWER
The inverse variation equation is y =
When x = –2, y =
28
= –14.
–2
28
x
What are the differences between direct,
inverse and joint variation?
~ y varies directly with x if y=ax for nonzero
constant a
~ y varies inversely with x if xy = a for a nonzero
constant a
~ z varies jointly with x and y if z = axy for
nonzero constant a


Y varied directly with x. If y = 36 when
x = 8, find y when x = 5




What are the steps for multiplying
& dividing rational expressions?


 Simplified
form of a rational expression ~ A
rational expression in which the numerator
and denominator have no common factors
other than 1 and -1
 Reciprocal
~ The multiplicative inverse of any
nonzero number
Simplify a rational expression
EXAMPLE 1
Simplify :
x2 – 2x – 15
x2 – 9
SOLUTION
(x +3)(x –5)
x2 – 2x – 15
=
(x +3)(x –3)
x2 – 9
=
(x +3)(x –5)
(x +3)(x –3)
x–5
=
x–3
ANSWER
x–5
x–3
Factor numerator and denominator.
Divide out common factor.
Simplified form
Standardized Test Practice
EXAMPLE 3
SOLUTION
8x 3y
2x y2
7x4y3 56x7y4
=
4y
8xy3
=
Multiply numerators
and denominators.
8 7 x
8
x6
y3 y
x y3
= 7x6y
ANSWER
The correct answer is B.
Factor and divide out
common factors.
Simplified form
EXAMPLE 4
Multiply:
Multiply rational expressions
3x –3x2
x2 + 4x – 5
x2 + x – 20
3x
SOLUTION
=
3x –3x2
x2 + 4x – 5
x2 + x – 20
3x
3x(1– x)
(x –1)(x +5)
(x + 5)(x – 4)
3x
Factor numerators and
denominators.
=
3x(1– x)(x + 5)(x – 4)
(x –1)(x + 5)(3x)
Multiply numerators and
denominators.
=
3x(–1)(x – 1)(x + 5)(x – 4)
(x – 1)(x + 5)(3x)
Rewrite 1– x as (– 1)(x – 1).
=
3x(–1)(x – 1)(x + 5)(x – 4)
(x – 1)(x + 5)(3x)
Divide out common factors.
EXAMPLE 4
Multiply rational expressions
= (–1)(x – 4)
Simplify.
= –x + 4
Multiply.
ANSWER
–x + 4
EXAMPLE 5
Multiply:
Multiply a rational expression by a polynomial
x+2
x3 – 27
(x2 + 3x + 9)
SOLUTION
x+2
x3 – 27
x+2
= 3
x – 27
=
=
=
(x2 + 3x + 9)
x2 + 3x + 9
1
(x + 2)(x2 + 3x + 9)
(x – 3)(x2 + 3x + 9)
(x + 2)(x2 + 3x + 9)
(x –
3)(x2
+ 3x + 9)
x+2
x–3
ANSWER
Write polynomial as a
rational expression.
Factor denominator.
Divide out common factors.
Simplified form
x+2
x–3
for Examples 3, 4 and 5
GUIDED PRACTICE
Multiply the expressions. Simplify the result.
8.
3x5 y2
8xy
6xy2
9x3y
SOLUTION
3x5 y2
2xy
6xy2
9x3y
=
=
=
18x6y4
Multiply numerators
and denominators.
72x4y2
18 x4
y2 x2
18 4
y2
x4 y2
Factor and divide out
common factors.
x2y2
4
Simplified form
for Examples 3, 4 and 5
GUIDED PRACTICE
9.
2x2 – 10x
x2– 25
x+3
2x2
SOLUTION
=
2x2 – 10x
x2– 25
x+3
2x2
2x(x –5)
(x –5)(x +5)
x+3
2x (x)
Factor numerators and
denominators.
=
2x(x –5) (x + 3)
(x –5)(x + 5)2x (x)
Multiply numerators and
denominators.
=
2x(x –5) (x + 3)
(x –5)(x + 5)2x (x)
Divide out common factors.
=
x+3
x(x + 5)
Simplified form
GUIDED PRACTICE
10.
x+5
x3– 1
for Examples 3, 4 and 5
x2 +x + 1
SOLUTION
x+5
x3– 1
=
=
=
=
x2 +x + 1
x+5
(x – 1)
(x2
+x + 1)
(x + 5) (x2 +x + 1)
(x – 1)
(x2
+x + 1)
(x + 5) (x2 +x + 1)
(x – 1) (x2 +x + 1)
x+5
x–1
x2 +x + 1
Factor denominators.
1
Multiply numerators and
denominators.
Divide out common factors.
Simplified form
EXAMPLE 6
Divide :
Divide rational expressions
7x
2x – 10
x2 – 6x
x2 – 11x + 30
SOLUTION
7x
2x – 10
x2 – 6x
x2 – 11x + 30
7x
=
2x – 10
x2 – 11x + 30
x2 – 6x
(x – 5)(x – 6)
=
7x
2(x – 5)
x(x – 6)
7x(x – 5)(x – 6)
=
2(x – 5)(x)(x – 6)
=
7
2
ANSWER
Multiply by reciprocal.
Factor.
Divide out common factors.
Simplified form
7
2
Divide a rational expression by a polynomial
EXAMPLE 7
6x2 + x – 15
4x2
Divide :
(3x2 + 5x)
SOLUTION
6x2 + x – 15
4x2
6x2 + x – 15
=
4x2
(3x + 5)(2x – 3)
=
4x2
=
=
(3x2 + 5x)
1
3x2 + 5x
1
x(3x + 5)
(3x + 5)(2x – 3)
4x2(x)(3x
Factor.
Divide out common factors.
+ 5)
2x – 3
4x3
ANSWER
Multiply by reciprocal.
Simplified form
2x – 3
4x3
for Examples 6 and 7
GUIDED PRACTICE
Divide the expressions. Simplify the result.
11.
x2 – 2x
x2 – 6x + 8
4x
5x – 20
SOLUTION
4x
5x – 20
x2 – 2x
x2 – 6x + 8
4x
=
5x – 20
x2 – 6x + 8
x2 – 2x
=
=
=
4(x)(x – 4)(x – 2)
5(x – 4)(x)(x – 2)
4(x)(x – 4)(x – 2)
5(x – 4)(x)(x – 2)
4
5
Multiply by reciprocal.
Factor.
Divide out common factors.
Simplified form
GUIDED PRACTICE
2x2 + 3x – 5
6x
12.
for Examples 6 and 7
(2x2 + 5x)
SOLUTION
2x2 + 3x – 5
6x
=
=
=
=
2x2 + 3x – 5
6x
(2x2 + 5x)
1
(2x2 + 5x)
Multiply by reciprocal.
(2x + 5)(x – 1)
6x(x)(2 x + 5)
(2x + 5)(x – 1)
6x(x)(2 x + 5)
x–1
6x2
Factor.
Divide out common factors.
Simplified form
What are the steps for multiplying & dividing
rational expressions?
Multiply: multiply the numerators /
multiply the denominators then simplify
Divide: multiply the first expression by the
reciprocal of thesecond
 expression, then
follow the rules for multiplication
Find the least common multiple of 20
and 45.




What are the steps for adding or
subtracting rational expressions
with different denominators?


 Complex
fraction ~ A fraction that contains a
fraction in its numerator or denominator.
Add or subtract with like denominators
EXAMPLE 1
Perform the indicated operation.
a.
7
3
+
4x
4x
b.
2x
x+6
–
5
x+6
SOLUTION
a.
7
3
+
4x
4x
b.
2x
x+6
–
7+3
=
4x
5
x+6
=
=
10
5
=
4x
2x
2x – 5
x+6
Add numerators and
simplify result.
Subtract numerators.
for Example 1
GUIDED PRACTICE
Perform the indicated operation and simplify.
a.
7–5
7
5
+
=
12x
12x
12x
=
2
1
=
12x
6x
b.
2+1
2
1
+
=
3x2
3x2
3x2
=
3
3x2
c.
4x–x
4x
x
–
=
x–2
x–2 x–2
=
=
1
x2
3x
3x
=
x–2
3x – 2
2+2
2(x2+1)
4x
2
2x
d.
+ 2
=
=
x2+1
x +1
x2+1
x2+1
=2
Subtract numerators
and simplify results .
Add numerators and
simplify results.
Subtract numerators.
Factor numerators and
simplify results .
EXAMPLE 2
Find a least common multiple (LCM)
Find the least common multiple of 4x2 –16 and
6x2 –24x + 24.
SOLUTION
STEP 1
Factor each polynomial. Write numerical factors as
products of primes.
4x2 – 16 = 4(x2 – 4) = (22)(x + 2)(x – 2)
6x2 – 24x + 24 = 6(x2 – 4x + 4) = (2)(3)(x – 2)2
EXAMPLE 2
Find a least common multiple (LCM)
STEP 2
Form the LCM by writing each factor to the highest power it occurs in
either polynomial.
LCM = (22)(3)(x + 2)(x – 2)2 = 12(x + 2)(x – 2)2
EXAMPLE 3
Add:
Add with unlike denominators
7
x
+
9x2
3x2 + 3x
SOLUTION
To find the LCD, factor each denominator and write each factor to the
highest power it occurs. Note that 9x2 = 32x2 and 3x2 + 3x = 3x(x + 1), so the
LCD is 32x2 (x + 1) = 9x2(x 1 1).
7
x
+
9x2 3x2 + 3x
7
9x2
x+1
x+1
=
+
7
9x2
x
3x(x + 1)
+
x
3x(x + 1)
3x
3x
Factor second denominator.
LCD is 9x2(x + 1).
EXAMPLE 3
Add with unlike denominators
3x2
+ 2
9x (x + 1)
=
7x + 7
9x2(x + 1)
=
3x2 + 7x + 7
9x2(x + 1)
Multiply.
Add numerators.
Subtract with unlike denominators
EXAMPLE 4
x+2
2x – 2
Subtract:
–
–2x –1
x2 – 4x + 3
SOLUTION
x+2
2x – 2
–
–2x –1
x2 – 4x + 3
=
x+2
– 2x – 1
–
(x – 1)(x – 3)
2(x – 1)
=
x+2
2(x – 1)
x–3
x–3
x2 – x – 6
=
2(x – 1)(x – 3)
–
–
– 2x – 1
(x – 1)(x – 3)
– 4x – 2
2(x – 1)(x – 3)
Factor denominators.
2
2
LCD is 2(x  1)(x  3).
Multiply.
EXAMPLE 4
=
x2 – x – 6 – (– 4x – 2)
2(x – 1)(x – 3)
x2 + 3x – 4
=
2(x – 1)(x – 3)
=
=
Subtract with unlike denominators
(x –1)(x + 4)
2(x – 1)(x – 3)
x+4
2(x –3)
Subtract numerators.
Simplify numerator.
Factor numerator.
Divide out common
factor.
Simplify.
GUIDED PRACTICE
for Examples 2, 3 and 4
Find the least common multiple of the polynomials.
5. 5x3 and 10x2–15x
STEP 1
Factor each polynomial. Write numerical factors as
products of primes.
5x3 = 5(x) (x2)
10x2 – 15x = 5(x) (2x – 3)
STEP 2
Form the LCM by writing each factor to the highest power it occurs in
either polynomial.
LCM = 5x3 (2x – 3)
for Examples 2, 3 and 4
GUIDED PRACTICE
Find the least common multiple of the polynomials.
6. 8x – 16 and 12x2 + 12x – 72
STEP 1
Factor each polynomial. Write numerical factors as
products of primes.
8x – 16 =
8(x – 2) =
12x2 + 12x – 72 =
23(x – 2)
12(x2 + x – 6) =
4 3(x – 2 )(x + 3)
STEP 2
Form the LCM by writing each factor to the highest power it occurs in
either polynomial.
LCM = 8 3(x – 2)(x + 3)
= 24(x – 2)(x + 3)
for Examples 2, 3 and 4
GUIDED PRACTICE
7.
3
4x
1
7
–
SOLUTION
3
4x
1
7
–
3
4x
7
7
–
21 – 4x
7(4x)
4x(7)
=
21 – 4x
28x
1
7
4x
4x
LCD is 28x
Multiply
Simplify
for Examples 2, 3 and 4
GUIDED PRACTICE
8.
x
1
+
3x2
9x2 – 12x
SOLUTION
x
1
+
3x2
9x2 – 12x
=
1
x
+
3x2
3x(3x – 4)
=
1
3x2
3x – 4
x
+
3x – 4
3x(3x – 4)
3x – 4
= 2
3x (3x – 4)
x2
+ 2
3x (3x – 4 )
Factor denominators
x
x
LCD is 3x2 (3x – 4)
Multiply
GUIDED PRACTICE
3x – 4 + x2
3x2 (3x – 4)
x2 + 3x – 4
3x2 (3x – 4)
for Examples 2, 3 and 4
Add numerators
Simplify
for Examples 2, 3 and 4
GUIDED PRACTICE
9.
x
x2 – x – 12
+
5
12x – 48
x
x2 – x – 12
+
5
12x – 48
x
(x+3)(x – 4)
+
5
12 (x – 4)
SOLUTION
=
x
=
(x + 3)(x – 4)
12x
=
12(x + 3)(x – 4)
12
12
+
5
+
12(x – 4)
Factor denominators
x+3
x+3
LCD is 12(x – 4) (x+3)
5(x + 3)
12(x + 3)(x – 4)
Multiply
GUIDED PRACTICE
for Examples 2, 3 and 4
=
12x + 5x + 15
12(x + 3)(x – 4)
Add numerators
=
17x + 15
12(x +3)(x + 4)
Simplify
for Examples 2, 3 and 4
GUIDED PRACTICE
10.
x+1
x2 + 4x + 4
6
x2 – 4
–
SOLUTION
x+1
x2 + 4x + 4
=
=
=
6
x2 – 4
–
x+1
(x + 2)(x + 2)
x+1
(x + 2)(x + 2)
x2 – 2x + x – 2
(x + 2)(x + 2)(x – 2)
–
6
Factor
denominators
(x – 2)(x + 2)
x–2
x–2
–
–
6
(x – 2)(x + 2)
x+2
x+2
LCD is (x – 2) (x+2)2
6x + 12
(x – 2)(x + 2)(x + 2)
Multiply
for Example 2, 3 and 4
GUIDED PRACTICE
=
x2 – 2x + x – 2 – (6x + 12)
(x +
x2 – 7x – 14
=
(x + 2)2 (x – 2)
2)2(x
Subtract numerators
– 4)
Simplify
Simplify a complex fraction (Method 2)
EXAMPLE 6
Simplify:
5
x+4
1
2
+
x+4
x
SOLUTION
The LCD of all the fractions in the numerator and
denominator is x(x + 4).
5
5
x+4
x+4
=
1
2
1
2
+
+
x+4
x
x+4
x
=
5x
x + 2(x + 4)
=
5x
3x + 8
x(x+4)
x(x+4)
Multiply numerator and
denominator by the LCD.
Simplify.
Simplify.
for Examples 5 and 6
GUIDED PRACTICE
11.
x
x
–
3
6
x
7
–
5
10
x
x
–
3
6
x
7
–
5
10
=
=
x
–
6
x
3
x
7
–
5
10
– 5x
3 (2x – 7)
30
30
Multiply numberator and
denominator by the LCD
Simplify
for Examples 5 and 6
GUIDED PRACTICE
2 –4
x
2
+3
x
2 –4
x
2
+3
x
=
=
=
2 – 4x
2 + 3x
x
x
12.
2 –4
x
2
+3
x
Multiply numberator and
denominator by the LCD
Simplify
2 (1 – 2x )
2 + 3x
Simplify
for Examples 5 and 6
GUIDED PRACTICE
3
x+5
13.
2
1
+
x–3
x+5
3
x+5
2
1
+
x–3
x+5
=
=
=
3
x+5
2
1
+
x–3
x+5
3x – 3
3x + 7
3(x – 3)
3x + 7
(x + 5)(x – 3)
(x + 5)(x – 3)
Multiply
numberator and
denominator by
the LCD
Simplify
Simplify
What are the steps for adding or subtracting
rational expressions with different
denominators?
1. Find the least common denominator
2. Rewrite each fraction using the common
denominator
3. Add or subtract
4. Simplify


Solve the
4
equation:
𝑥


=
6
15


What are the steps for solving
rational equations?


 Cross
mulitplying ~ A method for solving a
simple rational equation for which each side of
the equation is a single rational expression
 Extraneous
solution
Solve a rational equation by cross multiplying
EXAMPLE 1
9
3
=
4x + 1
x+1
9
3
=
4x + 1
x+1
Solve:
Write original equation.
3(4x + 5) = 9(x + 1)
Cross multiply.
12x + 15 = 9x + 9
Distributive property
3x + 15 = 9
3x = – 6
x=–2
Subtract 9x from each side.
Subtract 15 from each side.
Divide each side by 3.
ANSWER
The solution is –2. Check this in the original equation.
Standardized Test Practice
EXAMPLE 3
SOLUTION
7
5
9
+
= –
x
x
4
4x
(
)
Write original equation.
9
7
5
+
= 4x –
x
x
4
Multiply each side by the LCD, 4x.
20 + 7x = –36
Simplify.
7x = – 56
x=–8
Subtract 20 from each side.
Divide each side by 7.
EXAMPLE 3
Standardized Test Practice
ANSWER
The correct answer is B.
Solve a rational equation with two solutions
EXAMPLE 4
Solve:
1–
8
=
x–5
3
x
8
3
=
x–5
x
x(x – 5)
1– 8
=
x–5
1–
x(x – 5)
(
)
Write original equation.
3 Multiply each side by the LCD, x(x–5).
x
x(x –5) – 8x = 3(x – 5)
Simplify.
x2 – 5x – 8x = 3x – 15
Simplify.
x2 – 16x +15 = 0
(x – 1)(x – 15) = 0
x = 1 or x = 15
Write in standard form.
Factor.
Zero product property
EXAMPLE 4
Solve a rational equation with two solutions
ANSWER
The solutions are 1 and 15. Check these in the original
equation.
Check for extraneous solutions
EXAMPLE 5
Solve:
6
8x2
= 2
x–3
x –9
–
4x
x+3
SOLUTION
Write each denominator in factored form.
The LCD is (x + 3)(x – 3).
8x2
6
=
x –3
(x + 3)(x – 3)
(x + 3)(x – 3)
6
= (x + 3)(x – 3)
x –3
6(x + 3) = 8x2 – 4x(x – 3)
6x + 18 = 8x2 – 4x2 + 12x
–
4x
x+3
8x2
(x + 3)(x – 3)
(x + 3)(x – 3)
4x
x+3
EXAMPLE 5
Check for extraneous solutions
0 = 4x2 + 6x –18
0 = 2x2 + 3x – 9
0 = (2x – 3)(x + 3)
2x – 3 = 0 or x + 3 = 0
x=
3 or x = –3
2
You can use algebra or a graph to check whether either of the two
solutions is extraneous.
Algebra
3
The solution checks,
but the
2 apparent solution –3 is extraneous,
because substituting it in the equation results in division by zero, which
is undefined.
GUIDED PRACTICE
for Examples 3, 4 and 5
Solve the equation by using the LCD. Check for extraneous solutions.
5.
3
=3
x
7
+
2
SOLUTION
Write each denominator in factored form. The LCD is 2x
7
+
2
2x
7
+ 2x
2
3
=3
x
3
= 2x 3
x
7x + 6 = 6x
x= –6
for Examples 3, 4 and 5
GUIDED PRACTICE
6.
4
2
=2
+
3
x
SOLUTION
Write each denominator in factored form. The LCD is 3x
4
2
=2
+
3
x
3x
2
+ 3x
x
4
= 3x 2
3
6 + 4x = 6x
6 = 2x
x=3
for Examples 3, 4 and 5
GUIDED PRACTICE
7.
8
3
=1
+
x
7
SOLUTION
Write each denominator in factored form. The LCD is 7x
8
3
=1
+
x
7
7x
3
+ 7x
7
8
= 7x 1
x
3x + 56 = 7x
56 = 4x
x = 14
GUIDED PRACTICE
for Examples 3, 4 and 5
8. 3 + 4 = x +1
x –1
x –1
2
SOLUTION
Write each denominator in factored form. The LCD is 2( x – 1)
3
x +1
+ 4 =
x –1
x –1
2
(x – 1 )(2)
3
+ (x – 1)(2)
2
3x – 3 + 8= 2x + 2
x=–3
4
= (x – 1)(2)
x –1
x+1
x1
GUIDED PRACTICE
9.
3x
x +1
for Examples 3, 4 and 5
– 5 = 3
2x
2x
SOLUTION
Write each denominator in factored form. The LCD is (x + 1)(2x)
3x
x +1
2x (x + 1)
– 5 = 3
2x
2x
–3x2x (x +1)
x +1
= 2x (x5+1)
2x
3
2x
for Examples 3, 4 and 5
GUIDED PRACTICE
6x2 – 5x – 5 = 3x + 3
0 = 3x + 3 – 6x2 +5x + 5
0 = – 6x2 + 8x + 8
0 = (3x +2) (x – 2)
3x + 2 = 0
x=
–
2
3
or
x–2=0
x=2
GUIDED PRACTICE
10.
5x
x –2
= 7 +
for Examples 3, 4 and 5
10
x –2
SOLUTION
Write each denominator in factored form. The LCD is x – 2
x–2
5x = 7 + 10
x –2
x –2
5x
= (x – 2) 7 + (x – 2)
x –2
5x = 7x – 14 + 10
4 = 2x
x=2
x=2 results in no solution.
10
x–2
What are the steps for solving
rational equations?
If a proportion = find the cross product
If not a proportion – find the product by
using LCD of each expression, simplify,
check for extraneous

 solutions