Download Example 2

Mrs. Lusignea
Room 651
Algebra 2 Honors
Welcome to Algebra 2 Honors
This packet is a review of Algebra 1 Concepts that are essential for student success in Algebra 2.
We will go through this packet quickly.
If you need additional help with the concepts, please ask!
Name_____________________________________________
1
Date_______________________________________________
Mrs. Lusignea
Room 651
Algebra 2 Honors
I.
Simplifying Polynomial Expressions
A.
Combining Like Terms
You can add or subtract terms that are considered “like”, or terms that have the same variable(s) with
the same exponent(s).
Example 1:
5𝑥 − 7𝑦 + 10𝑥 + 3𝑦
5𝑥 + 10𝑥 − 7𝑦 + 3𝑦
15𝑥 − 4𝑦
Example 2:
−8ℎ2 + 10ℎ3 − 12ℎ2 − 15ℎ3
−8ℎ2 − 12ℎ2 + 10ℎ3 − 15ℎ3
−20ℎ2 − 5ℎ3
B.
Combining Like Terms AND Applying the Distributive Property
If there is no operation between the number and the parenthesis, every term inside the parenthesis is
multiplied by the term outside of the parenthesis.
Example 1:
3(4𝑥 − 2) + 13𝑥
3(4𝑥) − 3(2) + 13𝑥
12𝑥 − 6 + 13𝑥
25𝑥 − 6
Example 2:
3(12𝑥 − 5) − 9(−7 + 10𝑥)
3(12𝑥) + 3(−5) − 9(−7) − 9(10𝑥)
36𝑥 − 15 + 63 − 90𝑥
36𝑥 − 90𝑥 − 15 + 63
−54𝑥 + 48
2
Mrs. Lusignea
Room 651
Algebra 2 Honors
Practice Set 1
Simplify.
1. 8𝑥 − 9𝑦 + 16𝑥 + 12𝑦
2. 14𝑦 + 22 − 15𝑦 2 + 23𝑦
Simplify. Show all work.
3. 5𝑛 − (3 − 4𝑛)
4. −2(11𝑏 − 3)
5. 10𝑞(16𝑥 + 11)
6. −(5𝑥 − 6)
7. 3(18𝑧 − 4𝑤) + 2(10𝑧 − 6𝑤)
8. (8𝑐 + 3) + 12(4𝑐 − 10)
9. 9(6𝑥 − 2) − 3(9𝑥 2 − 3)
10. −(𝑦 − 𝑥) + 6(5𝑥 + 7)
Note: we use the word simplify (not solve) when we are combining like terms or evaluating expressions.
3
Mrs. Lusignea
Room 651
Algebra 2 Honors
II.
Evaluating Expressions


Substitute given value into expression using parenthesis.
Simplify using the order of operations.
Example 1:
Evaluate 𝑚 + (3 − 𝑞)2 if m = 12 and q = ‒ 1
( 12) + (3 − (−1))2
12 + (4)2
12 + 16
28
Example 2:
Evaluate
𝑥 4 −3𝑤𝑦
𝑦 3 +2𝑤
if w = 4, x = ‒ 3, and y = 5
(−3)4 −3(4)(−5)
(−5)3 +2(4)
81−3(4)(−5)
−125+2(4)
81−(−60)
−125+8
141
−117
𝑜𝑟 −
47
39
Practice Set 2
Evaluate each expression if h = 4, j = -1 and k = 0.5. Show all work.
1.
4
ℎ2 𝑘 + ℎ(ℎ − 𝑘)
2.
𝑗 + (3 − ℎ)2
3.
𝑗 2 −3ℎ2 𝑘
𝑗 3 +2
Mrs. Lusignea
Room 651
Algebra 2 Honors
III.
Solving Equations


Isolate the variable by using inverse operations.
CHECK YOUR ANSWERS!
Example 1:
8𝑥 + 4 = 4𝑥 + 28
−4=
Check:
− 4
8(6) + 4 = 4(6) + 28
48 + 4 = 24 + 28
8𝑥 = 4𝑥 + 24
52 = 52 √
−4𝑥 = −4𝑥
4𝑥 = 24
4𝑥
4
=
24
4
𝑥=6
Example 2:
5(4𝑥 − 7) = 8𝑥 + 45 + 2𝑥
20𝑥 − 35 = 10𝑥 + 45
20𝑥 − 10𝑥 − 35 = 10𝑥 − 10𝑥 + 45
10𝑥 − 35 = 45
10𝑥 − 35 + 35 = 45 + 35
10𝑥 = 80
10𝑥
10
80
= 10
𝑥=8
5
Check:
5(4(8) − 7) = 8(8) + 45 + 2(8)
5(32 − 7) = 64 + 45 + 16
5(25) = 125
125 = 125 √
Mrs. Lusignea
Room 651
Algebra 2 Honors
Practice Set 3
Solve each equation. You must show all work.
1.
−131 = −5(3𝑥 − 8) + 6𝑥
2.
−7𝑥 − 10 = 18 + 3𝑥
3.
12𝑥 + 8 − 15 = −2(3𝑥 − 82)
4.
−(12𝑥 − 6) = 12𝑥 + 6
5.
3
(15𝑑
5
6.
8(1 + 7𝑥) + 6 = 14 − 5𝑥
7.
−(1 − 7𝑛) = −3 + 7𝑛
8.
−4(𝑏 + 7) = −28 − 4𝑏
6
1
+ 20) − 6 (18𝑑 − 12) = 38
Mrs. Lusignea
Room 651
Algebra 2 Honors
IV.
Solving Literal Equations


A literal equation is an equation that contains more than one variable.
You can solve a literal equation by isolating the specified variable.
Example 1:
3𝑥𝑦 = 18, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥
3𝑥𝑦
3𝑦
18
= 3𝑦
6
y≠0
𝑥 = 𝑦;
𝑑=
2𝑥
𝑎
+ 𝑏, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥
𝑑−𝑏 =
2𝑥
𝑎
+𝑏−𝑏
𝑑−𝑏 =
2𝑥
𝑎
Example 2:
2𝑥
𝑎
𝑎(𝑑 − 𝑏) = 𝑎 ( )
𝑎(𝑑 − 𝑏) = 2𝑥
𝑎(𝑑−𝑏)
2
=
𝑥=
2𝑥
2
𝑎(𝑑−𝑏)
;
2
a≠0
Practice Set 4
Solve for the specified variable. Show all work. Be sure to state restrictions.
1.
2𝑑 − 3𝑓 = 9,
3.
𝑃 = (𝑔 − 9)180,
7
for f
for g
2.
𝑑𝑥 + 𝑡 = 10,
4.
4𝑥 + 𝑦 − 5ℎ = 10𝑦 + 𝑢,
for x
for x
Mrs. Lusignea
Room 651
Algebra 2 Honors
V.
Absolute Value Expressions


Absolute value measures the distance from zero on the number line.
The absolute value of any number (other than zero) is positive.
Example:
Evaluate 8.4 − |2𝑛 + 5| if n = ‒ 7.5
8.4 − |2(−7.5) + 5|
8.4 − |−15 + 5|
8.4 − |−10|
8.4 − 10
−1.6
Practice Set 5
1.
Evaluate |4𝑥 + 3| − 3
1
2
if x = ‒ 2
2.
Evaluate 1 − |2𝑦 + 1| if 𝑦 = −
3.
Evaluate 2|−𝑥 + 7| − 25
if x = ̶ 3
4.
Evaluate |3𝑟 − 7 | if r = 0
8
1
3
2
3
Mrs. Lusignea
Room 651
Algebra 2 Honors
VI.
Absolute Value Equations




To solve an absolute value equation, first isolate the absolute value expression
In |𝑥 − 𝑐| = 𝑟
c is called the central value and r is called the range.
The absolute value expression is set equal to the range and it’s opposite.
Must check for extraneous solutions. An extraneous solution is a solution of an equation derived from an
original equation that is not a solution of the original equation.
|2𝑦 − 4| = 12
Example 1:
2𝑦 − 4 = 12
or
2𝑦 − 4 = −12
2𝑦 = 16
or
2𝑦 = −8
𝑦=8
or
𝑦 = −4
Check:
|2(8) − 4| = 12
|2(−4) − 4| = 12
|16 − 4| = 12
|−8 − 4| = 12
|12| = 12
|−12| = 12
12 = 12
12 = 12
3|4𝑤 − 1| − 5 = 10
Example 2:
3|4𝑤 − 1| = 15
|4𝑤 − 1| = 5
4𝑤 − 1 = 5
or
4𝑤 − 1 = −5
4𝑤 = 6
or
4𝑤 = −4
or
𝑤 = −1
3
𝑤=2
Check:
3
9
3 |4 (2) − 1| − 5 = 10
3|4(−1) − 1| − 5 = 10
3|6 − 1| − 5 = 10
3|−4 − 1| − 5 = 10
3|5| − 5 = 10
3|−5| − 5 = 10
15 − 5 = 10
15 − 5 = 10
Mrs. Lusignea
Room 651
Algebra 2 Honors
Practice Set 6
Solve the following. Check for extraneous solutions.
1.
2|3𝑥 − 1| + 5 = 33
2.
|2𝑥 + 3| = 3𝑥 + 2
3.
|𝑥| = 𝑥 − 1
4.
|4 − 𝑧| − 10 = 1
10