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Solving Equations
MODEL for SOLVING EQUATIONS
Solve the equation 3x + 2 = x + 4:
Step 1: Set up the equation.
Step 2: Remove (subtract)
two blue pawns from each side
of the scale.
Step 3: Remove (subtract) a
value of 2 from each side of
the scale.
2
2
2
4
4
2
Step 4: A blue pawn has a
value of 2.
2
2
4
-
=
x=2
Step 5: Check your solution by substituting 2 back into the original set up.
2
=
4
2+2+2+2
=
2+2+4
x = 2 is the correction solution.
8=8
Students simultaneously draw the model (from above) and solve algebraically:
Step 1:
+2
Step 2:
××
+2
××
+2
-2
+4
Step 4:
3x  2  2 x  4
2 x  2 x
+4
-2
Step 3:
Algebraically:
+4
+2
1x  2 
2
4
2

2
x
Step 5: Check your solution.
+2
2+2+2+2
=
+4
= 2+2+4
8=8
3 2  2  2  2  4
62  44
8=8
2
Solving and Graph Inequalities (Models)
Solving and Graphing INEQUALITIES Models
Example 1:
Solve the inequality x + 9 < 5.
Step 1: Set up the inequality.
Step 2: Remove (subtract) a
value of 9 from each side of the
scale.
Step 3: Each pawn has a value
less than -4.
9
9
<
<
<
Step 4: Check your
solution by substituting a
value for the pawn that is
less than -4.
Since -7 is less than -4,
let’s substitute it for
our blue pawn.
9
<
-4
-7 + 9
x < -4
<
5
< 5
2<5
x < -4
is the correction solution.
Example 2:
Solve the inequality 2x < 5.
Step 1: Set up the inequality.
>
Step 2: Divide 6 evenly among
each blue pawn.
Step 3: Each pawn has a value
greater than 3.
>
>
Step 4: Check your
solution by substituting a
value for each pawn that
is greater than 3.
Since 5 is greater than
3, let’s substitute it for
our blue pawns.
>
>
x>3
3
6
5+5 > 6
10 > 6
x>3
is the correction solution.
Graphing Inequalities on a Number Line
The following are examples of graphing the inequalities <, ≤, >, and ≥ on a number line.
0
0
x< 7
x£ - 2
0
0
x> 4
x³ - 4
Solve and graph the following inequalities.
1.
2b  4  10
4
2b
2
4
 14
2
b  7
-7
0
2.
12  3b  3
3
3.
b  6  2
6 6
3
b
15  3b
3
3
1
5b
0
 4 divide by –,
1
change the
sign
b  4
5
-4
0
In solving Inequalities, explain why dividing by a negative reverses the direction of the inequality sign.
Students often struggle with the concept that division by a negative reverses the direction of the inequality sign. Let’s look of some
examples to show why this works.
Example 1:
We know that 2 is most certainly less than 3, hence the statement 2 < 3 is a true statement. But if you multiply the inequality by
-1, you get -2 < -3. This inequality is saying that -2 smaller than -3, but this is a FALSE statement.
Any positive number gets larger as the absolute value of that positive number gets larger, but any negative number gets SMALLER as
the absolute value of that number gets LARGER. This makes working with inequalities and multiplying by negatives tricky. Think
about it in terms of sets of numbers, for example x ≥ -5 (see below). This is the set of all numbers that are larger than -5.
Now let’s explore with this statement. Suppose I want to add 3 to both sides of this inequality: x + 3≥ -5 +3 or x + 3 ≥ -2, will the
same set of numbers as above make this statement true? Yes. If I substitute -4 in for x the statement will be true, but if I substitute -6
it will not. I can do the same thing buy trying to subtract 3 from both sides, or multiplying by any positive number. But let’s try
simply multiplying both sides by -1: -x ≥ 5 would be the new statement, is this statement still true for the values above, let’s again
check -4 and -6. If I substitute -4 into the new inequality, I get 4 ≥ 5 NOT true, this is a contradiction to the original inequality. If I
substitute -6, I get a true statement which is also a contradiction.
Example 2:
When you multiply or divide by a negative number you have to reverse the inequality. Why?
Well, just look at the number line!
For example, from 3 to 7 is an increase, but from -3 to -7 is a decrease. See how the inequality sign reverses (from < to >)?
-7
-3
3
-7 < -3
7
7>3
Solve: -2y < -8
Divide both sides by -2 ... and reverse the inequality!
2 y  8
2
2
y4
And that is the correct solution: y > 4
(Note that I reversed the inequality on the same line I divided by the negative number.)
So, just remember:
When multiplying or dividing by a negative number, reverse the inequality
Problem Solving with Inequalities:
1. Alex earns $7.50 per hour by working after school. Alex needs at least $60 to buy a
video game. Write and solve an inequality that shows how many hours he must work
to buy the video game.
7.50 x  60
x  8 hours
7.50
7.50
2. To make a profit, Elwin’s Electronics has to sell more than 200 items in a month. If
Elwin sold 30 items in the first week of the month, write and solve an inequality that
shows how many items need to be sold in the remaining weeks to earn a profit.
x  30  200
 30  30
x
 170 items
Use Algebraic Properties to Justify Steps in Solving Equations
Algebra Properties: Let a, b, and c be real numbers, variables, or algebraic
expressions.
Property
Commutative Property of Addition
Example
a+b=b+a
3x + x2 = x2 + 3x
(3 - x)x2 = x2(3 - x)
Commutative Property of Multiplication ab = ba
Associative Property of Addition
(a + b) + c = a + (b
+ c)
(x + 3) + x2 = x + (3 + x2)
Associative Property of Multiplication
(ab)c = a(bc)
(3x • 2)(5) = (3x)(2 • 5)
a(b + c) = ab + ac
3x(5 + 2x) = (3x• 5) + (3x • 2x)
(a + b)c = ac + bc
(y + 5)4 = (y • 4) + (5 • 4)
Additive Identity Property
a+0=a
7x2 + 0 = 7x2
Multiplicative Identity Property
a•1=a
8y • 1 = 8y
Additive Inverse Property
a + (-a) = 0
5x2 + (-5x2) = 0
Multiplicative Inverse Property
1
a• =1
a
(x2 + 3) •
Symmetric Property of Equality
If a = b then b= a
Exponential Property of Equality
If b is a positive number other than 1, then bx = by if and only if
x = y.
Distributive Properties
1
=1
(x + 3)
2
Sample problem: Justify the steps in solving an equation:
(4 x  3)  7  6 x  6
Given
4 x  (3  7)  6 x  6
Associative property of addition
Subtract the same amount from both sides
4 x  10  6 x  6
4x
4x
of an equation. Maintain equality in an
10  2 x  6
equation
10  2 x  6
Subtract the same amount from both sides
of an equation. Maintain equality in an
6
6
equation
4  2x
Divide both sides of an equation by the same
4 2x

number. Maintain equality in an equation.
2 2
2 x
x 2
Symmetry property of Equality
EXTENSION Solving Absolute Value and Compound Inequalities
Solving Absolute Value and Compound inequalities
Solving an absolute value inequality is similar to solving compound inequalities.
It is important that students understand the meaning of absolute value and how it effects the solving of equations. Rather than
starting with rules/algorithms, it is helpful to start by developing an understanding of the meaning of absolute value equations
and then linking the meaning to the algorithmic process.
|x| = 5 means, some value “x” is exactly 5 units from the origin. There are two such values for x, 5 and -5. It is helpful for
students to see this on a number line.
Note that the origin is in the middle and that the values for x are each 5 units left or right of the origin. Notice also that the
“gap” between the solutions is 10 or twice 5 because we are looking for the values of x that are exactly 5 units from the origin.
Now suppose we want to find |x| > 5. This means all values of 5 that are more than 5 units from the origin. Building on the
previous, let’s start by graphing.
From the graph above, we can see that on the right all values to the right of 5 are more than 5 units from the origin. For
example, 6, 7, 8…and all values in between are more than 5 units from the origin. To express these values, we can write x > 5.
Now let’s turn our attention to the region on the left. There, all values left of -5 are more than 5 units from the origin. Here,
values such as -6,-7,-8…are more than 5 units from the origin. Writing these values may seem counter intuitive to students
because we are saying that they are more than 5 units from the origin, but to express them mathematically we say x < -5. A
conversation about this is with students essential. Thus, the solution to |x| > 5 is x > 5 or x < -5. Any value for x that meets
either of those conditions, satisfies |x| > 5. Again notice that the “gap” is 10 units.
ABSOLUTE VALUE Equations and Inequalities (continued)
Now suppose we want to find |x| < 5. Again, starting with the meaning is essential. The equation |x| > 5 means we are looking
for all values that are less than 5 units from the origin. As with the previous examples, let’s start with a picture.
From the graph above, we can see that we need to include all values for x that are between -5 and 5. There is a significant
difference in this problem situation over the previous one. In the previous example, any value to the right of 5 or to the left of -5
satisfied the inequality. In this situation, we cannot say any value to the left of 5 or to the right of -5 will satisfy the in equality.
For example, if we say any value to the right of -5 (x > -5) then x=10 would satisfy that inequality, but it does NOT satisfy |x|<5.
Rather, we need all the values that are BOTH to the right of -5 (x > -5) AND to the left of 5 (x < 5). There are two ways this is
typically written: a) x > -5 and x < 5, b) -5< x <5. Again the “gap” is 10 units.
After exploring simple cases as above, student should begin to realize that when solving an absolute value inequality, there are
always two “parts”. From here, teachers want to develop the following:
For |x| ≥ a or |x| > a we are looking for distances greater than a, hence we will have x > a OR x < -a
For |x| ≤ a or |x| < a we are looking for distances less than a, we are looking for –a < x < a
|x| ≥ a
|x| ≤ a
2a
2a
-a
a
x ≤ -a
x≥a
x ≥ a OR x ≤ -a
-a
a
x ≥ -a
x≤a
-a ≤ x ≤ a
ABSOLUTE VALUE / COMPOUND INEQUALITIES (continued)
Example 1: (solving with “greater than”)
Example 2: (solving with “less than or equal to”)
Solve: |x – 20| > 5
Case 1:
x – 20 > 5
x > 25
or
Case 2:
x – 20 < -5
x < 15
x < 15 or x > 25
Note that there are two parts to the solution and that the
connecting word is “or”. Also note that 15 and 25 are 2x5 or
10 units apart
Example 3: (isolating the absolute value first)
Solve: |3 + x| – 4 < 0
Case 1:
|3 + x| < 4
3+x<4
x<1
and
Case 2:
|3 + x| < 4
3 + x > -4
x > -7
Solve: |x – 3| ≤ 4
Case 1:
x–3≤ 4
x≤7
and
Case 2:
x – 3 ≥ -4
x ≥ -1
x ≥ -1 and x ≤ 7
also written as:
-1 ≤ x ≤ 7
Note that there are two parts to the solution and that the
connecting word is “and”. Al.so note that -1 and 7 are 2x4 or
8 units apart
Example 4: (all values work)
WHOA!!
Don’t solve
this problem!
Solve: |x + 4| > -3
Case 1:
x + 4 > -3
x > -7
or
Case 2:
x+4<3
x < -1
x < 1 and x > -7
also written as:
x > -7 or x < -1
-7 < x < 1
Answer: x Є 
Note that the absolute value is isolated before the solution
begins.
You already know the answer! Absolute value is ALWAYS
positive (or zero) so it is always > -3. All values work.
ABSOLUTE VALUE / COMPOUND INEQUALITIES (continued)
Example 5: (no values work)
Example 6: (word problem)
WHOA!!
Don’t solve
this problem!
Solve: |x + 1| < -6
Case 1:
x + 1 < -6
x > -7
and
Case 2:
x+1>6
x>5
x < -7 and x > 5 ???
Answer: 
(the empty set)
You already know the answer! Absolute value is ALWAYS
positive (or zero). It is NEVER < -6. No values work!
At Brooks Graphic Company, the average starting salary
for a new graphic designer is $37, 600, but the actual
salary could differ from the average by as much as $2590.
a.
b.
Write an absolute value inequality to describe
this situation.
Solve the inequality to find the range of the
starting salaries.
Solution:
a. |x – 37,600| ≤ 2590
|the difference between the average and the salary| ≤
$2590
b. The absolute value |x – a| < b represents the set of all
points x that are less than b units away from a.
Case 1:
|x – 37,600| ≤ 2590
x – 37,600 ≤ 2590
x ≤ 40190
and
Case 2:
|x – 37,600| ≤ 2590
x – 37,600 ≥ -2590
x ≥ 35010
$35,010 ≤ x ≤ $40,190
Set and Interval Notation
Set Notation: Set notation is mathematical language for precisely stating all numbers of a specific set that possess a specific
property.
real
numbers;
 integer
 natural numbers
numbers;
Interval notation: An interval is a connected subset of numbers. Interval notation is an alternative to expressing your answer
as an inequality. Unless specified otherwise, it is assumed that we are indicating real numbers, when working with interval
notation. This notation is probably the least familiar to students.
When using interval notation, the
symbol:
( means “not included” or “open”.
[
2≤x<6
means “included” or “closed”.
[2, 6)
as an inequality.
in interval notation.
Interval Notation:
description
Open Interval: (a, b) is interpreted as a < x < b where the endpoints are
NOT included. (While this notation resembles an ordered pair, in this context
it refers to the interval upon which you are working.)
diagram
(1, 5)
[1, 5]
Closed Interval: [a, b] is interpreted as a < x < b where the endpoints are
included.
(1, 5]
Half-Open Interval: (a, b] is interpreted as a < x < b where a is not included,
but b is included.
[1, 5)
Half-Open Interval: [a, b) is interpreted as a < x < b where a is included, but
b is not included.
Non-ending Interval: (a, ∞) is interpreted as x > a where a is not included
and infinity is always expressed as being "open" (not included).
(1, ∞)
(-∞, 5]
Non-ending Interval: (-∞, b] is interpreted as x < b where b is included and
again, infinity is always expressed as being "open" (not included).
For some intervals it is necessary to use combinations of interval notations to achieve the desired set of numbers.
Consider how you would express the interval "all numbers except 13".
As an inequality:
In interval notation:
x < 13 or x > 13
(-∞, 13)
(13, ∞)
Notice that the word "or" has been replaced with the symbol "U",
which stands for "union".
Consider expressing in interval notation, the set of numbers which contains all numbers less than 0 and also all
numbers greater than 2 but less than or equal to 10.
As an inequality:
x < 0 or 2 < x < 10
In interval notation:
(-∞, 0)
(2, 10]
As you have seen, there are many ways of representing the same interval of values. These ways may include word
descriptions or mathematical symbols. The following statements and symbols ALL represent the same interval.
WORDS:
SYMBOLS:
"all numbers between positive
one and positive five, including
the one and the five."
"x is less than or equal to 5 and
greater than or equal to 1"
1<x<5
{x 
"x is less than or equal to 5 and
greater than or equal to 1"
Interval notation example: [ means “included” or “closed”.
“open”.
So, 2  x  6 in interval notation would be [2, 6).
Inequalities: Graph, Set Notation, Interval Notation
Inequality or Set
1.
Graph
| 1  x  5}
[1, 5]
( means “not included” or
Interval Notation
1  x  2
(1,2)
{-4, -3, -2, -1, 0, 2, 3}
4..3
2.
3.
x  2or x  5
4.
,..., 2, 1 or 7,8,..., 
( ..  2,  1] (6, 7.. )
5.
x 2
(  ,  2]
(, 2]
(5, )
Help Links
Solving Equations:
 http://www.purplemath.com/modules/solvelin.htm
 http://regentsprep.org/Regents/math/ALGEBRA/AE2/LSolvEq.htm
 Video: http://www.khanacademy.org/math/algebra/solving-linearequations/v/solving-equations-1
Write/solve word-problems:
 http://www.homeschoolmath.net/teaching/teach-solve-word-problems.php
Practice:
 http://hotmath.com/help/gt/genericalg1/section_4_2.html
Set and interval notation:
 http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntervalNot.htm
Practice interval notation:
 http://www.regentsprep.org/Regents/math/ALGEBRA/AP1/IntPrac.htm