Transcript x - Nutley Public Schools

ONE STEP
EQUATIONS
ONE STEP EQUATIONS
An equation is like a balance scale
because it shows that two
quantities are equal.
What you do to one side of the
equation must also be done to the
other side to keep it balanced.
ONE STEP EQUATIONS
To solve one step equations, you need to
ask three questions about the equation:
• What is the variable?
• What operation is performed on the variable?
• What is the inverse operation?
(The one that will undo what is being
done to the variable)
ONE STEP EQUATIONS
Example 1 Solve x + 4 = 12
What is the variable? The variable is x.
What operation is being performed on the variable?
Addition.
What is the inverse operation (the one that will undo what is being
done to the variable)?
Subtraction.
Using the subtraction property of equality, subtract 4 from both sides
of the equation.
The subtraction property of equality tells us to subtract
the same thing on both sides to keep the equation equal.
x + 4 = 12
-4
-4
x = 8
ONE STEP EQUATIONS
Example 2 Solve y - 7 = -13
What is the variable?
The variable is y.
What operation is being performed on the variable?
Subtraction.
What is the inverse operation (the one that will undo what is being done to
the variable)?
Addition.
Using the addition property of equality, add 7 to both sides of the
equation.
The addition property of equality tells us to add the same
thing on both sides to keep the equation equal.
y - 7 = -13
+7 +7
y = -6
ONE STEP EQUATIONS
Example 3 Solve –6a = 12
What is the variable?
The variable is a.
What operation is being performed on the variable?
Multiplication.
What is the inverse operation (the one that will undo what is being done to the
variable)?
Division
Using the division property of equality, divide both sides of the
equation by –6.
The division property of equality tells us to divide the same
thing on both sides to keep the equation equal.
–6a = 12
-6
-6
a = -2
ONE STEP EQUATIONS
Example 4 Solve b = -10
2 The variable is b.
What is the variable?
What operation is being performed on the variable?
Division.
What is the inverse operation (the one that will undo what is being done
to the variable)?
Multiplication
Using the multiplication property of equality, multiply both sides of
the equation by 2.
The multiplication property of equality tells us to multiply
the same thing on both sides to keep the equation equal.
b
= -10
2
b
2 •
= -10 • 2
2
b = -20
Solving Two-Step Equations
What is a Two-Step Equation?
An equation written in the form
Ax + B = C
Examples of Two-Step Equations
a) 3x – 5 = 16
b) y/4 + 3 = 12
c) 5n + 4 = 6
d) n/2 – 6 = 4
Steps for Solving
Two-Step Equations
1.
Solve for any Addition or Subtraction on the
variable side of equation by “undoing” the
operation from both sides of the equation.
2.
Solve any Multiplication or Division from variable
side of equation by “undoing” the operation from
both sides of the equation.
Opposite Operations
Addition  Subtraction
Multiplication  Division
Helpful Hints?
Identify what operations are on the
variable side. (Add, Sub, Mult, Div)
“Undo” the operation by using
opposite operations.
Whatever you do to one side, you
must do to the other side to keep
equation balanced.
Ex. 1: Solve 4x – 5 = 11
4x – 5 =
+5
4x
=
4
x=5
15
+5 (Add 5 to both sides)
20 (Simplify)
4 (Divide both sides by 4)
(Simplify)
Try These Examples
1. 2x – 5 = 17
2. 3y + 7 = 25
3. 5n – 2 = 38
4. 12b + 4 = 28
Check your answers!!!
1. x = 11
2. y = 6
3. n = 8
4. b = 2
Ready to Move on?
Ex. 2: Solve x/3 + 4 = 9
x/3 + 4 = 9
-4 -4
x/3 = 5
(Subt. 4 from both sides)
(Simplify)
(x/3)  3 = 5  3 (Mult. by 3 on both sides)
x = 15
(Simplify)
Try these examples!
1. x/5 – 3 = 8
2. c/7 + 4 = 9
3. r/3 – 6 = 2
4. d/9 + 4 = 5
Check your answers!!!
1. x = 55
2. c = 35
3. r = 24
4. d = 9
Time to Review!
•
•
•
Make sure your equation is in the form Ax + B = C
Keep the equation balanced.
Use opposite operations to “undo”
•
Follow the rules:
1. Undo Addition or Subtaction
2. Undo Multiplication or Division
Before we begin…
• In previous lessons we solved simple one-step linear
equations…
• In this lesson we will look at solving multi-step
equations…as the name suggests there are steps you
must do before you can solve the equation…
• To be successful here you have to be able to analyze
the equation and decide what steps must be taken to
solve the equation…
Multi-Step Equations
• This lesson will look at:
–
–
–
–
–
Solving linear equations with 2 operations
Combining like terms first
Using the distributive property
Distributing a negative
Multiplying by a reciprocal first
• Your ability to be organized and lay out your problem
will enable you to be successful with these concepts
Linear Equations with 2 Functions
• Sometimes linear equations will have more
than one operation
• In this instance operations are defined as
add, subtract, multiply or divide
• The rule is
– First you add or subtract
– Second you multiply or divide.
Linear Equations with 2 Operations
• You will have to analyze the equation first
to see what it is saying…
• Then, based upon the rules, undo each of
the operations…
• The best way to explain this is by looking at
an example….
Example # 1
2x + 6 = 16
In this case the problem says 2 times a number plus 6 is equal to
16
Times means to multiply and plus means to add
The first step is to add or subtract.
To undo the addition of 6 I have to subtract 6 from both sides
which looks like this:
16 minus
2x + 6 = 16
6 equals
The 6’s
- 6 -6
10
on the left
2x
= 10
side
cancel
What is left is a one step
each other
equation 2x = 10
out
Example #1 (Continued)
2x = 10
The second step is to multiply or divide
To undo the multiplication here you would divide both sides
by 2 which looks like this:
10 divided
2x = 10
The 2’s on
by 2 is
2
2
the left
equal to 5
cancel out
leaving x
x =5
The solution to the equation 2x + 6 = 16 is x = 5
Combining Like Terms First
• When solving multi-step equations, sometimes you have
to combine like terms first.
• The rule for combining like terms is that the terms must
have the same variable and the same exponent.
Example:
• You can combine x + 5x to get 6x
• You cannot combine x + 2x2 because the terms do not
have the same exponent
Example # 2
7x – 3x – 8 = 24
I begin working on the left side of the equation
On the left side I notice that I have two like terms (7x, -3x)
since the terms are alike I can combine them to get 4x.
7x – 3x – 8 = 24
4x – 8 = 24
After I combine the terms I have a 2-step equation. To solve
this equation add/subtract 1st and then multiply/divide
Example # 2 (continued)
4x – 8 = 24
Step 1: Add/Subtract
Since this equation has – 8, I will add 8 to both
sides
4x – 8 = 24
+8 +8
The 8’s
on the
left
cancel
out
4x
= 32
I am left with a 1-step
equation
24 + 8 = 32
Example # 2 (continued)
4x
= 32
Step 2: Multiply/Divide
In this instance 4x means 4 times x. To undo the
multiplication divide both sides by 4
32  4 = 8
4x
= 32
The 4’s on
4
4
the left
x
=8
cancel out
leaving x
The solution that makes the
statement true is x = 8
Solving equations using the Distributive
Property
• When solving equations, sometimes you will need to
use the distributive property first.
• At this level you are required to be able to recognize
and know how to use the distributive property
• Essentially, you multiply what’s on the outside of the
parenthesis with EACH term on the inside of the
parenthesis
• Let’s see what that looks like…
Example #3
5x + 3(x +4) = 28
In this instance I begin on the left side of the equation
I recognize the distributive property as 3(x +4). I must simplify
that before I can do anything else
5x + 3(x +4) = 28
5x +3x +12 = 28
After I do the distributive property I see that I have like terms
(5x and 3x) I have to combine them to get 8x before I can
solve this equation
Example # 3(continued)
8x + 12 = 28
I am now left with a 2-step equation
Step 1: Add/Subtract
The left side has +12. To undo the +12, I subtract 12 from both
sides
8x + 12 = 28
28 – 12 = 16
The 12’s
-12 -12
on the left
8x
= 16
cancel out
leaving 8x
Example # 3 (continued)
8x
= 16
Step 2: Multiply/Divide
On the left side 8x means 8 times x. To undo the multiplication
I divide both sides by 8
8x
The 8’s on
the left
cancel out
leaving x
= 16
8
8
x
=2
The solution that makes the statement true is x = 2
16  8 = 2
Distributing a Negative
• Distributing a negative number is similar to
using the distributive property.
• However, students get this wrong because
they forget to use the rules of integers
• Quickly the rules are…when multiplying, if
the signs are the same the answer is positive.
If the signs are different the answer is
negative
Example #4
4x – 3(x – 2) = 21
I begin by working on the left side of the equation.
In this problem I have to use the distributive property.
However, the 3 in front of the parenthesis is a negative 3.
When multiplying here, multiply the -3 by both terms within
the parenthesis. Use the rules of integers
4x – 3(x – 2) = 21
4x – 3x + 6 = 21
After doing the distributive property, I see that I can combine
the 4x and the -3x to get 1x or x
Example # 4 (continued)
4x – 3x + 6 = 21
x + 6 = 21
After combining like terms you are left with a simple one step
equation. To undo the +6 subtract 6 from both sides of the equation
The 6’s
cancel
out
leaving
x
x
+ 6 = 21
-6
-6
x
= 15
The solution is x = 15
21 – 6 = 15
Multiplying by a Reciprocal First
• Sometimes when doing the distributive property
involving fractions you can multiply by the reciprocal
first.
• Recall that the reciprocal is the inverse of the fraction
and when multiplied their product is equal to 1.
• The thing about using the reciprocal is that you have to
multiply both sides of the equation by the reciprocal.
• Let’s see what that looks like…
Example # 5
12 
3
( x  2)
10
In this example you could distribute the 3/10 to the x and the
2.
The quicker way to handle this is to use the reciprocal of 3/10
which is 10/3 and multiply both sides of the equation by 10/3
12
10 I F
10 I 3
F
 GJ ( x  2)
G
J
H3 K H3 K10
On the left side of the equation, after multiplying by the
reciprocal 10/3 you are left with 120/3 which can be simplified
On
the right side of the equation the reciprocals cancel each
to 40
other out leaving x + 2
The new equation is:
40 = x + 2
Example #5 (continued)
40 = x + 2
After using the reciprocals you are left with a simple one-step
equation
To solve this equation begin by working on the right side
and subtract 2 from both sides of the equation
40 = x + 2
The 2’s on
40 – 2 = 38
the right
-2
-2
side
38 = x
cancel out
leaving x
The solution to the equation is x = 38
Comments
• On the next couple of slides are some practice
problems…The answers are on the last slide…
• Do the practice and then check your answers…If you do
not get the same answer you must question what you
did…go back and problem solve to find the error…
• If you cannot find the error bring your work to me and I
will help…
Your Turn
1.
2.
3.
4.
5.
6x – 4(9 –x) = 106
2x + 7 = 15
6 = 14 – 2x
3(x – 2) = 18
12(2 – x) = 6
Your Turn
b g
6.
9
x  3  27
2
7.
4
 2 x  4  48
9
8.
5m – (4m – 1) = -12
9.
55x – 3(9x + 12) = -64
10.
9x – 5(3x – 12) = 30
b g
Your Turn Solutions
1.
2.
3.
4.
5.
14.2
4
4
8
1½
6. 3
7. -52
8. -13
9. -1
10. 5
Warm Up
Solve each equation.
1. 2x – 5 = –17
–6
2.
14
Solve each inequality and graph the
solutions.
3. 5 < t + 9
4.
t > –4
a ≤ –8
Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
1. Add/ Subtract
45 + 2b > 61
45 + 2b > 61
–45
–45
2b > 16
2. Multiply/Divide
The solution set is {b:b > 8}.
b>8
0
2
4
6
8
10 12 14
16 18 20
Solving Multi-Step Inequalities
Solve the inequality and graph the solutions.
1. Add/Subtract
8 – 3y ≥ 29
8 – 3y ≥ 29
–8
–8
2. Multiply/Divide
–3y ≥ 21
y ≤ –7
The solution set is {y:y  –7}.
–7
–10
–8 –6 –4 –2
0
2
4
6
8 10
You try
Solve and Graph the inequality
1) - 9 < 2b -13
2)2x + 13 > 9
y
3) + 12 £ 8
5
4) - 3m + 5 < 17
To solve more complicated inequalities, you may
first need to simplify the expressions on one or
both sides.
Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
2 – (–10) > –4t
1. Combine like terms.
12 > –4t
2. Multiply/Divide
–3 < t
(or t > –3)
The solution set is {t:t > –3}.
–3
–10
–8 –6 –4
–2
0
2
4
6
8 10
Simplifying Before Solving Inequalities
Solve the inequality and graph the solutions.
–4(2 – x) ≤ 8
1. Distributive Property
−4(2 – x) ≤ 8
−4(2) − 4(−x) ≤ 8
–8 + 4x ≤ 8
+8
+8
4x ≤ 16
2. Add/Subtract
3. Multiply/Divide
x≤4
–10
–8 –6 –4
–2
The solution set is {x:x ≤ 4}.
0
2
4
6
8 10
Now you try…
1. 3x – 7 > 2
x>3
4. x – 4 < 3
5
x < 35
2. 4x + 1  -3
x ≥ -1
3. 2x – 7 ≤ -3
x≤2
5. 15 + x ≥ 6
3
x ≥ -27
Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. 13 – 2x ≥ 21
x ≤ –4
2. –11 + 2 < 3p
p > –3
3. 23 < –2(3 – t)
t>7
4.
3.3 Absolute Value Equations and Inequalities
Summary:
Solving Absolute Value Equations and Inequalities
Let k be a positive real number, and p and q be real numbers.
1. To solve |ax + b| = k, solve the following compound equation.
ax + b = k
or
ax + b = –k.
The solution set is usually of the form {p, q}, which includes two
numbers.
p
Copyright © 2010
Pearson Education, Inc.
q
Sec 3.3 - 56
3.3 Absolute Value Equations and Inequalities
Summary:
Solving Absolute Value Equations and Inequalities
Let k be a positive real number, and p and q be real numbers.
2. To solve |ax + b| > k, solve the following compound inequality.
ax + b > k
or
ax + b < –k.
The solution set is of the form (-∞, p) U (q, ∞), which consists of two
separate intervals.
p
Copyright © 2010
Pearson Education, Inc.
q
Sec 3.3 - 57
3.3 Absolute Value Equations and Inequalities
Summary:
Solving Absolute Value Equations and Inequalities
Let k be a positive real number, and p and q be real numbers.
3. To solve |ax + b| < k, solve the three-part inequality
–k < ax + b < k
The solution set is of the form (p, q), a single interval.
p
Copyright © 2010
Pearson Education, Inc.
q
Sec 3.3 - 58
3.3 Absolute Value Equations and Inequalities
EXAMPLE
1
Solve |2x + 3| = 5.
Solving an Absolute Value Equation
3.3 Absolute Value Equations and Inequalities
EXAMPLE
2
Solve |2x + 3| > 5.
Solving an Absolute Value Inequality with >
3.3 Absolute Value Equations and Inequalities
EXAMPLE
3
Solve |2x + 3| < 5.
Solving an Absolute Value Inequality with <
3.3 Absolute Value Equations and Inequalities
EXAMPLE
4
Solving an Absolute Value Equation That
Requires Rewriting
Solve the equation |x – 7| + 6 = 9.
3.3 Absolute Value Equations and Inequalities
Special Cases for Absolute Value
Special Cases for Absolute Value
1.
2.
The absolute value of an expression can never be negative: |a| ≥ 0
for all real numbers a.
The absolute value of an expression equals 0 only when the
expression is equal to 0.
Copyright © 2010
Pearson Education, Inc.
Sec 3.3 - 63
3.3 Absolute Value Equations and Inequalities
EXAMPLE
6
Solving Special Cases of Absolute Value
Equations
Solve each equation.
(a)
|2n + 3| = –7
See Case 1 in the preceding slide. Since the absolute value of an
expression can never be negative, there are no solutions for this equation.
The solution set is Ø.
(b)
|6w – 1| = 0
See Case 2 in the preceding slide. The absolute value of the expression 6w – 1 will equal 0 only if
6w – 1 = 0.
The solution of this equation is 1 . Thus, the solution set of the original
6
equation is { 1 }, with just one element. Check by substitution.
6
Copyright © 2010
Pearson Education, Inc.
Sec 3.3 - 64
3.3 Absolute Value Equations and Inequalities
EXAMPLE
7
Solving Special Cases of Absolute Value
Inequalities
Solve each inequality.
(a)
|x| ≥ –2
The absolute value of a number is always greater than or equal to 0.
Thus, |x| ≥ –2 is true for all real numbers. The solution set is (–∞, ∞).
(b)
|x + 5| – 1 < –8
Add 1 to each side to get the absolute value expression alone on one
side.
|x + 5| < –7
There is no number whose absolute value is less than –7, so this inequality
has no solution. The solution set is Ø.
Copyright © 2010
Pearson Education, Inc.
Sec 3.3 - 65
3.3 Absolute Value Equations and Inequalities
EXAMPLE
7
Solving Special Cases of Absolute Value
Inequalities
Solve each inequality.
(c)
|x – 9| + 2 ≤ 2
Subtracting 2 from each side gives
|x – 9| ≤ 0
The value of |x – 9| will never be less than 0. However, |x – 9| will equal 0
when x = 9. Therefore, the solution set is {9}.
Copyright © 2010
Pearson Education, Inc.
Sec 3.3 - 66
Solving an Absolute-Value Inequality
Solve | x  4 | < 3
x  4 IS POSITIVE
|x4|3
x  4  3
x7
x  4 IS NEGATIVE
|x4|3
x  4  3
x1
Reverse
inequality symbol.
The solution is all real numbers greater than 1 and less than 7.
This can be written as 1  x  7.
Solving an Absolute-Value Inequality
Solve
 1POSITIVE
| 3  6 and graph
2x +| 2x
1 IS
2x +the1solution.
IS NEGATIVE
| 2x  1 |  3  6
2x + 1 IS POSITIVE
| 2x|2x
1|
 31 |6 9
2x 1 |1 
| 2x
9 +9
| 2x  1 | 3  6
2x + 1 IS NEGATIVE
| 2x|2x
1|
31 |6  9
2x

1

9
| 2x  1 |  9
2x

10

2x  1 9
x  5
2x  8
x4
2x  1  +9
2x  10
2x  8
The solution is all real numbers greater than or equal
x4
x  5
to 4 or less than or equal Reverse
to  5. This can be written as
the compound inequality inequality
x   5 or x  4.
symbol.
6 5 4 3 2 1
0
1
2
3
4
5
6
Strange Results
2(3x  8)  7  5
(2[3 x  (8  4)]  12) 3  2
True for All Real Numbers,
since absolute value is
always positive, and
therefore greater than any
negative.
No Solution Ø.
Positive numbers are
never less than
negative numbers.
Examples
2w  5  7
2w  5  7 or 2w  5  7
2w  2
2w  12
w  1
or
w6
Check and verify on a number line. Numbers above 6 or below
-1 keep the absolute value greater than 7. Numbers between
them make the absolute value less than 7.
Key Skills
Solve absolute-value inequalities.
Solve |x – 4|  5.
Case 1:
Case 2:
x – 4 is positive
x – 4 is negative
x–45
x – 4  –5
x9
x  –1
solution: –1  x  9
Key Skills
Solve absolute-value inequalities.
Solve |4x – 2|  -18.
Exception
alert!!!!
When the absolute value
equals a negative value,
there is no solution.
Key Skills
TRY THIS
Solve absolute-value inequalities.
Solve |2x – 6|  18.
Case 1:
2x – 6 is positive
2x – 6  18
2x  24
x  12
Solution: –6  x  12
Case 2:
2x – 6 is negative
2x - 6  –18
2x  –12
x  –6
Key Skills
TRY THIS
Solve absolute-value inequalities.
Solve |3x – 2|  -4.
Exception
alert!!!!
When the absolute value
equals a negative value,
there is no solution.