Download Chapter 7 Class Notes Intermediate Algebra, MAT1033C SI Leader Joe Brownlee

Chapter 7 Class Notes
Intermediate Algebra, MAT1033C
SI Leader Joe Brownlee
Palm Beach State College
Class Notes
7.1
Professor Burkett
SI Leader Joe Brownlee
7.1 – Rational Expressions and Functions; Multiplying and Dividing
Chapter 7 takes factoring to the next level, so make sure you’re set on the basics!
In 7.1 we’re dealing with rational expressions and functions, which is a fancy way of saying fractions.
Example 1:
Find all numbers that are not in the domain.
𝑓 (𝑥 ) =
𝑥+4
𝑥−6
When we’re asked to find the numbers that are not in the domain, all we do is set the denominator equal to
zero and solve.
𝑥−6= 0
𝑥 = 6 Add six to both sides.
𝑥 ≠ 6 x cannot equal six.
The reason x cannot equal six is because if it did, it would make the denominator 0 (6-6=0); as we know, the
denominator cannot equal 0, otherwise it is undefined.
Example 2:
Find all numbers that are not in the domain.
𝑓 (𝑥 ) =
3 + 2𝑥
5
In this case, since there is no variable in the denominator, nothing we plug-in for x will make the denominator 0. There
Example 3:
Find all numbers that are not in the domain.
𝑓 (𝑥 ) =
2
𝑥2 − 𝑥 − 6
Factor the denominator into (𝑥 + 2)(𝑥 − 3) and set each factor equal to zero.
𝑥+2= 0
𝑥 = −2
𝑥−3= 0
𝑥=3
Therefore, x cannot equal -2 or 3, otherwise the denominator will equal 0, which will make it undefined.
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Class Notes
7.1
Professor Burkett
SI Leader Joe Brownlee
Next we’ll be writing rational expressions in lowest terms. Let’s practice a couple easier problems before
tackling more difficult problems.
Example 4:
Write each rational function in lowest terms.
8𝑘
16
We see that the 8 and 16 can be reduced:
8𝑘
16 2
𝑘
So we’re left with 2.
Example 5:
Write the rational function in lowest terms.
𝑦+2
𝑦2 − 4
𝑦+2
Factor the denominator (difference of two squares).
(𝑦+2)(𝑦−2)
𝑦+2
Cancel.
(𝑦+2)(𝑦−2)
1
= (𝑦−2)
Lowest terms.
Example 6:
𝑦−2
2−𝑦
When the numerator and denominator are exact opposites, the rational expression is reduced to -1.
= -1
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Class Notes
7.1
Professor Burkett
SI Leader Joe Brownlee
Example 7:
𝑚2 − 16
1
∙
𝑚+2 𝑚+4
(𝑚+4)(𝑚−4)
1
∙
Factor each numerator and denominator into lowest terms.
𝑚+2
𝑚+4
(𝑚+4)(𝑚−4)
1
∙
Cancel like terms.
𝑚+2
𝑚+4
(𝑚−4) 1
∙ Multiply numerators, multiply denominators.
𝑚+2 1
=
𝑚−4
𝑚+2
Example 8:
𝑦 2 − 2𝑦 − 3
𝑦2 − 1
÷
𝑦 2 + 4𝑦 + 4 𝑦 2 + 𝑦 − 2
When dividing fractions, we flip the second fraction and turn the division sign into multiplication. Sometimes
this is referred to as “keep, change, flip.”
𝑦 2 − 2𝑦 − 3 𝑦 2 + 𝑦 − 2
∙
𝑦 2 + 4𝑦 + 4
𝑦2 − 1
(𝑦+1)(𝑦−3) (𝑦+2)(𝑦−1)
∙
Factor each numerator and denominator into lowest terms.
(𝑦+2)(𝑦+2) (𝑦+1)(𝑦−1)
(𝑦+1)(𝑦−3) (𝑦+2)(𝑦−1)
(𝑦−3)
∙
Cancel.  = (𝑦+2)
(𝑦+2)(𝑦+2) (𝑦+1)(𝑦−1)
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Class Notes
7.2
Professor Burkett
SI Leader Joe Brownlee
7.2 – Adding and Subtracting Rational Expressions
When adding and subtracting fractions, they must have the same (or “common”) denominator. If they start
with the same denominator, great; just deal with the numerators. If the denominators are different, you must
make them the same, by multiplying as necessary, then deal with the numerators.
Example 1: Basic Review
Example 2:
Notice the denominator was the same in start, so we rewrite the
problem with the numerators combined over one denominator.
Remove the parenthesis in the numerator, ensuring to distribute
the negative.
Combine like terms in the numerator.
Factor the GCF out of the numerator and cancel.
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Class Notes
7.2
Professor Burkett
SI Leader Joe Brownlee
Example 3:
Notice the denominators in the original are different. We must not
find a common denominator in order to continue. In this case, the
common denominator will be each of the original denominators,
(x+1) (x-4).
Once we find the common denominator, we multiply the numerator by what
8
was missing in the original denominator. In this case, the 𝑥+1 was missing the
(x-4) in its denominator, so when we put it in the denominator, we must
multiply the numerator by that same binomial, because whatever we do to
the top we must also do to the bottom. Therefore, we get 8(x-4) in the
numerator.
5
Repeat this process for the 𝑥−4 to get 5(x+1) in the new numerator.
Example 4:
Rewrite the fractions, factoring both denominators.
Determine the common denominator, then multiply
the numerators as necessary.
Now that the denominators are equal, combine and simplify
the numerator.
In this case, when we simplified the numerator, nothing could
cancel.
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Class Notes
7.4
Professor Burkett
SI Leader Joe Brownlee
7.4 – Equations with Rational Expressions and Graphs
In this section, we’ll build on the skills learned in 7.1 and 7.2.
Example 1:
Solve the equation.
3 1
5
+ =
𝑥 2 6𝑥
3(6𝑥) 1(6𝑥) 5(6𝑥)
+
=
𝑥
2
6𝑥
18 + 3𝑥 = 5
Find your common denominator and multiply each numerator by it to clear your
fractions.
Clear your fractions and simplify.
3𝑥 = −15
𝑥 = −5
Solve for x.
Example 2:
Solve the equation.
1
𝑥
−8
+
= 2
𝑥 + 4 𝑥 − 4 𝑥 − 16
1
𝑥
−8
+
=
𝑥 + 4 𝑥 − 4 (𝑥 + 4)(𝑥 − 4)
Factor the denominators to better determine a common denominator.
1(𝑥 − 4)(𝑥 + 4) 𝑥(𝑥 + 4)(𝑥 − 4) −8(𝑥 + 4)(𝑥 − 4)
+
=
(𝑥 + 4)(𝑥 − 4)
𝑥+4
𝑥−4
𝑥 − 4 + 𝑥 2 + 4𝑥 = −8
1
Multiply each fraction by the common
denominator to clear the fractions.
Class Notes
7.4
𝑥 2 + 5𝑥 + 4 = 0
Combine like terms and set the equation equal to zero.
(𝑥 + 1)(𝑥 + 4) = 0
Factor.
𝑥+1=0
𝑥+4=0
Professor Burkett
SI Leader Joe Brownlee
Set each factor to 0.
𝑥 = −4, −1
𝑥 = −1
You must check to see if plugging the solution into the original equation will make one of the
denominators 0. In this case, if we plugged -4 into the original equation, we get a zero for two
of the denominators. Therefore, we must drop that as an answer and only use -1.
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Class Notes
7.5
Professor Burkett
SI Leader Joe Brownlee
7.5 – Applications of Rational Expressions
Example 1:
Solve for p, if f=15 and q=25.
1 1 1
= +
𝑓 𝑝 𝑞
1
1 1
= +
15 𝑝 25
Plug in the given values where indicated.
1(75𝑝) 1(75𝑝) 1(75𝑝)
=
+
15
𝑝
25
5𝑝 = 75 + 3𝑝
Multiply each fraction by the common denominator to clear the fractions.
Simplify.
2𝑝 = 75
𝑝=
75
2
Solve for p.
Example 2:
Solve for the specified variable.
𝐴=
𝑅𝑟
𝑓𝑜𝑟 𝑅.
𝑅+𝑟
𝐴(𝑅 + 𝑟) =
𝑅𝑟(𝑅 + 𝑟)
𝑅+𝑟
𝐴𝑅 + 𝐴𝑟 = 𝑅𝑟
𝐴𝑟 = 𝑅𝑟 − 𝐴𝑅
Multiply both sides of the equation by the common denominator to clear the fraction.
Simplify.
Since we’re solving for R, bring all terms with R to one side of the equation.
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Class Notes
𝐴𝑟 = 𝑅(𝑟 − 𝐴)
𝑅=
7.5
Professor Burkett
SI Leader Joe Brownlee
Factor out R from the right side. Then divide to get R by itself.
𝐴𝑟
𝑟−𝐴
Example 3:
Lauren’s car uses 15 gal. of gasoline to drive 495 miles. She has 6 gal. of gas in the car, and she wants to know
how much more gasoline she will need to drive 600 miles.
15 𝑔𝑎𝑙𝑙𝑜𝑛𝑠 𝑥 𝑔𝑎𝑙𝑙𝑜𝑛𝑠
=
495 𝑚𝑖𝑙𝑒𝑠
600 𝑚𝑖𝑙𝑒𝑠
15(600) = 495𝑥
Set up a simple proportion.
Cross multiply.
9000 = 495𝑥
𝑥 = 18.2 𝑔𝑎𝑙𝑙𝑜𝑛𝑠
Simplify.
Therefore, if Lauren already has 6 gallons, she will need an additional 12.2 gallons of gas to make her 600 mile
journey.
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