Download Chapter 4

Chapter 4
Linear Equations and Inequalities in One
Variable
View the Video Tutorial for an overview of Linear Equations.
With the introduction of variables in algebra, we can now talk about equations which could not be done
with just arithmetic. An equation consists of two expressions, at least one of which contains a variable, set
equal to each other. To solve an equation is to nd all values that make the equation true. For example to
solve
x+3=5
means nd all numbers,
x,
such that if you add 3 to it, you get 5. As we will nd, in the following sections,
the only number that satises this condition is 2. So we would say that the solution is
x = 2.
It should be noted that our description of an equation is not the most general denition, but does describe
the type of equations that we will work with. There are many types and classications of equations which
depend on the number of variables in the equation and what type of operations are involved. The creation
of equations and the methods of solving them is the primary goal of mathematics. The ability to create and
solve equations allows us to answer a vast range of real life problems. Problems such as how much a person
should be saving each month if they wish to retire in 20 years, what will the weather be like tomorrow, to
how long will radio-active waste be dangerous, and what should be the holding pattern and descent of a
passenger jet plane to minimize passenger discomfort.
All the mathematical concepts that we learn are to help us solve various types of equations.
In this
chapter, we begin the exploration of equations by studying the most fundamental type of equation, the
linear equation in one variable. The concepts and techniques that you will learn in this chapter will be used
over and over again throughout the rest of your mathematical studies.
4.1
Inverses
To understand how we solve equations, we must rst understand the idea of inverses. In mathematics, an
inverse is something that will undo some operation. So, in talking about inverses, we need to also talk about
what operation we are undoing.
4.1.1
Additive Inverse
To understand what an additive inverse does, we begin with an example. Start with the number 3. If we
add 5 to it, we get 8. Now, if we add
Start with the
−5
to 8, we get back to the number that we started with, 3.
Add 5
Now add
−5
We get the number we
number
⇒
3
So adding the
started with.
−5
3+5=8
undoes the adding of 5.
⇒
8 + (−5)
This means that
we have that
1
−5
⇒
is the additive inverse of 5.
3
In general,
CHAPTER 4.
2
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Denition 4.1.1.
Additive Inverse
The additive inverse of a number is the opposite of the number. In other words, the
additive inverse of
Example 4.1.1.
1)
a
is
−a.
Find the additive inverse of the following:
2
3
2)
−3
3)
4x
Solutions:
2
3 is
1) The opposite of
− 23 .
−a is the additive
a, −a does not mean a negative number. −a just means the opposite of whatever a is. In
since a = −3, the opposite of −3 is 3, i.e.−a = −(−3) = 3. So the additive inverse of −3 is
2) Note that in the denition of additive inverse, Denition 4.1.1, when we say that
inverse of
this case,
3.
3) The opposite of
4.1.2
4x
−4x.
is
Multiplicative Inverse
Just as the additive inverse undoes addition, a multiplicative inverse would undo multiplication. We use the
following example to illustrate what the multiplicative inverse does.
Start with the
Multiply
number
by 5
⇒
3
Now multiply
by
3 · 5 = 15
⇒
15 ·
1
5 1
5
We get the number we
started with
⇒
3
1
1
So multiplying by
5 undoes the multiplication by 5, which means that 5 is the multiplicative inverse of
5. In general, we have
Denition 4.1.2.
Multiplicative Inverse
The multiplicative inverse of a number is its reciprocal. In other words, the multiplicative inverse of
•
a
1
a
Another way to say this is that the multiplicative inverse of
Example 4.1.2.
1)
is
b
a
b is a .
Find the multiplicative inverse.
2
3
2)
−3
3) 0
Solutions:
1) The multiplicative inverse of
2) The multiplicative inverse of
3
2
3 is its reciprocal 2 .
−3
is its reciprocal
of inverse we want. The additive inverse of
−3
− 31 .
Note that we need to pay attention to what type
is 3 but the multiplicative inverse is
− 13 .
1
0 , which is undened. This means that 0 has no multiplicative inverse. This
makes sense because if you multiply any number by 0 you get 0. For example 2 · 0 = 0 and 3 · 0 = 0.
3) The reciprocal of 0 is
Now if 0 had a multiplicative inverse, we should be able to undo the multiplication by 0 and get back
to the number before we multiplied by 0. But then the question would be what number to we go back
to? 2 or 3? There is no way to know and so we cannot undo the multiplication by 0, hence 0 has no
multiplicative inverse.
CHAPTER 4.
4.2
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
3
Addition Property of Equality
View the Video Tutorial for this section and the next section here.
To solve equations requires two concepts. The idea of undoing an operation, which we discussed with the
inverses (Additive and Multiplicative) and the idea of balance. The idea of balance is as follows. Suppose
we have
x=2
This means that
weight as
x
x
x
is the same as
2.
One way to visualize this is as follows. We label one box of unknown
and the other that weights 2 pounds with 2.
x=2
would mean that if we place the box labeled
on one side of the scale and the box labeled 2 on the other, then the scale would be balanced.
Now if we added 3 pounds to the left side (x), the scale would no longer be balanced.
But, if we added the 3 pounds to both sides of the scale, it would remain balanced.
So mathematically this means that if we start with
x=2
and add 3 to both sides of the equation, it would still be equal.
x+3
=
2+3
or
x+3
=
5
Now there are two implications to the above example. The rst is that if we add the same thing to both sides
of the equation, then they are still equal to each other. The second is a little more subtle. The solution(s)
CHAPTER 4.
4
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
to an equation, is the number(s) that make the equation true. So the solution to the equation
be 2, since if we replace the
x
x=2
would
with 2, we get
x
=
2
(2)
=
2
which is true. Notice that 2 is also the solution to the new equation we got since
x+3
=
5
(2) + 3
=
5
5
=
5
So the second implication is that if you add the same thing to both sides of an equation, then the old equation
and the new equation will have the same solutions.
If we generalize this, we get
Proposition 4.2.1.
Addition Property of Equality.
If
a=b
then
a+c=b+c
The addition property has two implications.
•
If you have an equation and you add the same thing to both sides of the equation,
then they are still equal.
•
The old equation and the new equation have the same solutions.
Now, if you have had any experience solving equations or read the beginning of this chapter, you may
be scratching your head because my example above seems to be going in reverse order. I started with
and got
x + 3 = 5.
x=2
You would be justied in your quandary. Normally, when we solve an equation, the goal
is to get the variable by itself on one side of the equation and everything else on the other side. Keep in
mind that the above example is of the addition property, Proposition 4.2.1, not about solving an equation.
We now turn our attention to solving equations using the addition property of equality and the inverses.
To solve the equation
right side
left side
z }| {
x+3 =
z}|{
5
as we stated before, we must try to get the variable, in this case
we need to get rid of the 3. Since the 3 is connected to
x
x,
by itself on one side of the equation. So
by addition, to undo the addition by 3, we need
−3. But according to the addition property of equality, Proposition 4.2.1, if we
−3 to the left side of the equation, then we must also add it to the right side to keep the equation balanced.
to add the additive inverse
add
(x + 3) + (−3) = 5 + (−3)
Add -3 to both sides of the equation. There are
parentheses around
x+3
because we are adding the
-3 to the entire side, which consists of both the
x
and
the 3.
x + 3 + (−3) = 2
However, recall that for addition, the parentheses do
nothing and can be ignored (note that this will not
be true when we talk about multiplication in the
next section.)
x+0=2
x=2
Note that we can check to see if this is the correct answer for the equation by plugging in 2 for any
x
we see
in the original equation and see if the two sides actually come out the same. So, the check is as follows:
CHAPTER 4.
5
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
x+3=5
(2) + 3 = 5
Plug in 2 for the
x.
5=5
Since this is true,
x=2
is the solution to the
equation.
Example 4.2.1.
Solve.
1)
x−7=9
3)
x + a − b = c,
2)
2
5
4)
x + 3 = 7 + 2x
+x=2
solve for
x.
Solutions:
1)
x−7=9
x + (−7) = 9
We rewrite the subtraction as adding the opposite.
Since -7 is connected to
x
by addition, we add 7 to
undo it. To balance the equation, we add the 7 to
x + (−7) =
9
both sides of the equation.
+7
+7
Notation: We write what we want to add to both
sides of the equation below the equation itself.
x = 16
2) Since
2
5 is added to
x,
we undo the addition by adding
− 23
to both sides of the equation.
Notation: expressions written below an equation
2
+x =
32
−3
0+x=
6
3
are understood to be added to the equation. Here,
22
we are adding
−3
+ − 23
− 32
To add fractions, they must have the same
denominator. So
x=
to both sides of the equation.
2=
2
1
3
3
6
3
4
3
a, b, and c
x+a−b=c
x + a + (−b) = c
3) We treat the variables
just as we would numbers.
Convert subtraction to addition by the opposite.
Add the additive inverse of
−a
x + a + (−b) =
−a
=
+b
and
b,
a
and
−b,
which are
to both sides of the equation.
c
−a+b
x=c−a+b
Since there are no like terms, there is nothing to
combine here.
4) The goal of solving an equation is to get the variable we want to solve for by itself on one side of the
equation and everything else on the other side. In this problem, notice that
equation. So we need to gather all the
x's
on one side of the equation.
x
is on both sides of the
CHAPTER 4.
6
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
To get rid of the
x
on the left side, since it is
added to 3 we add its opposite,
x +3
=
−x
3
−7
=
7 + 2x
−x
7 +x
−7
−4 = x
of the equation
−x
to both sides
CHAPTER 4.
4.3
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
7
Multiplication Property of Equality
View the Video Tutorial for this section and the revious section here.
As stated in the previous section, to solve equations requires two concepts.
operation (inverses) and the idea of balancing an equation.
The idea of undoing an
In the last section, we discussed how adding
the same quantity to both sides of an equation keeps the equation balanced. Here we will discuss how an
equation is balanced when multiplication is involved. As before, we begin with
x=2
which we can visualize as two boxes that balance a scale.
If we triple the left side, i.e. multiply
x
by 3, note that the scale would be o balance.
To balance the scale, we would have to multiply both sides by 3.
Which gives us
3x
=
3(2)
3x
=
6
As with the addition property of equality, there are two implications to the multiplication property. First,
that multiplying both sides of the equation by the same number or expression keeps the equation balanced
and that the new and old equations both have the same answers. With the multiplication property, there
are a couple of subtle points that we must be careful of that did not exist with the addition property. We
will discuss these after rst stating the multiplication property.
CHAPTER 4.
8
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Proposition 4.3.1.
Multiplication Property of Equality.
If
a=b
then
c(a) = c(b)
for any
c 6= 0.
The multiplication property has two implications.
•
If you have an equation and you multiply the same nonzero expression to both
sides of the equation, they are still equal.
•
The old equation and the new equation have the same solutions.
As we stated before, there are a couple of subtle but important points to keep in mind when using the
multiplication property of equality.
The rst is that it does not apply if you multiply both sides of the
equation by the number zero. The reason for this is that if you multiply any number by 0, you get 0. So a
false equation can be turned into a true statement. For example,
2 = −5
We know that 2 does not equal -5. So this equation is
never true.
0(2) = 0(−5)
0=0
Now multiply both sides by zero.
This equation is always true.
So we went from an equation,
2 = −5,
which is never true to an equation,
0 = 0,
which is always true.
These equations cannot be equivalent to each other and so we see that multiplying both sides of an equation
by zero can lead to misleading answers.
As to the second issue, it can be better discussed through an example. We start with the equation
x+3=5
Suppose now that we wish to double the right side, i.e. multiply the 5 by 2. The question is, what do
we do to the left side. Note that if we just double the
of 5 since it take both
In other words,
x
x
that will not be enough to balance out the doubling
and 2 to balance out 5.
2x + 3 6= 5.
To balance the equation, if we double the right side, then we must double
everything on the left side of the equation as well. In other words we must double the 3 as well as the
x.
CHAPTER 4.
9
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
In algebraic form, we mean that
x+3=5
Given this equation, if we double the right side, then
we must double the entire left side.
2(x + 3) = 2(5)
Notice that unlike the addition property, the
parentheses cannot be ignored when applying the
multiplication property of equality.
2x + 6 = 10
It should be noted here that the above example is to explain the nuances of the multiplication property
of equality and not an example of how to solve equations. The above is just a demonstration of how you
multiply both sides of the equation by a number and still maintain the equality. To solve equations using
the multiplication property, we need to choose carefully what we will multiply both sides of the equation by.
The choice of what to multiply by is determined by what multiplication we wish to undue, which involves
the use of multiplicative inverses.
Example 4.3.1.
Solve.
1)
5)
3)
2
− x=6
5
−3x = −5
2)
−x = 3
6)
4)
x
=4
5
−2x
=6
5
(a + 3)x = 5,
solve for
x
Solutions:
1) Remember that to solve an equation, we must get the variable, in this case
x,
by itself on one side of
the equation.
−3x = −5
Since the
−3
is multiplied to
x,
to get rid of it, we
need to multiply by the multiplicative inverse of
−3,
− 13 (−3x) = − 13 (−5)
− 31 .
1
multiply −
3
which is
If we
on one side, we must also
multiply it on the other side of the equation to
keep it balanced.
x=
2)
x
1
5 can be viewed as 5
·x
5
3
and so we multiply by 5 to undo the multiplication by
x
= 4
x 5
= 5(4)
5
5
x = 20
3) The multiplicative inverse of
− 25
is
− 25 ,
so
2
− x =
5
5
2
−
− x
=
2
5
x =
6
5
− (6)
2
−15
1
5 . So
CHAPTER 4.
10
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
−2x
2
5 is the same as − 5 x, so this equation is really the same equation as the previous example
and can of course be solved the same way. However, an alternate way to solve the equation would be
4) Note that
to view the
x
as being multiplied by
viewing it as one number
− 25
−2
and divided by 5 and so undo those two operations instead of
that is multiplied to
x.
−2x
=6
5
Since dividing by 5 really means multiply by
5
−2x
5
1
5 , we
undo it by multiplying by 5.
= 5(6)
Multiplicative inverse of -2 is
− 21 .
1
1
− (−2x) = − (30)
2
2
x = −15
5) Recall that
−x
can be viewed as
−1 · x.
Note that the reciprocal of -1 is
−x
=
1
−1
= −1.
So
3
−1(−x) = −1(3)
x = −3
6) Since we are solving for x, we need to get rid of the expression
(a + 3)
(a + 3)x
1
· (a + 3)x
a+3
=
x
=
Note here that we do not want to distribute the
wrong with distributing in the
x,
x
=
into
x
a + 3,
1
(5)
a+3
5
a+3
(a + 3).
Although there is nothing algebraically
x.
In fact, if we distribute the
in them. Since we are trying to get
of the equation, this would be counter productive.
which is
5
it would not help us to isolate the
then we end up with two terms that have
x. To
1
a+3 . So
which is multiplied to
undo this multiplication we need to multiply by the multiplicative inverse of
x
x,
by itself on one side
CHAPTER 4.
4.4
11
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Summary of Solving Linear Equations in One Variable
4.4.1
General Equations
View the Video Tutorial for this section here.
In general, the equations that we encounter will have a combination of addition, subtraction, multiplication and division within the equation, which will require us to use a combination of the addition and
multiplication properties of equality as well as the various inverses. The point of confusion for students often
comes down to what do you do rst, because there are often more than one way to solve an equation. Recall
that to solve an equation, we use the additive and multiplicative inverses to undo the various operations to
get the variable by itself. Usually, when you want to undo a sequence of events, you work in the reverse order.
For example, after you shower, the sequence that you follow to dress would be to put on your undergarments
rst, then your outer garments. To take a shower, you would have to undo what you did to get dressed. To
do this, you would start by removing your outer garments rst and then your undergarments. So to undress,
you reverse the sequence you followed to get dressed. Likewise, when we solve equations, we use the order
of operations in reverse.
To solve the equation
2x + 3 = 11
note that on the left side of the equation, if
x
had been a number and you were just evaluating
of operations would say that you rst multiply by 2 then add the 3. To solve for
x
2x + 3,
order
means we must undo the
multiplying by 2 and the addition of 3. To do this, we work in the reverse of the order of operations and
rst undo the addition by 3. So we add
−3
to both sides of the equation.
2x + 3
=
2x
=
−3
11
−3
8
Now, we undo the multiplication.
1
1
(2x) =
(8)
2
2
x = 4
Example 4.4.1.
Solve.
1)
4 − 6x = 8
3)
1 + 4(x − 3) = 3(3 − 2x) − 2x
2)
x − 3[x − (2 − x)] = −2
4)
a(x − 2) = bx − 3,
solve for
x
Solutions:
1) Note that
4 − 6x
can be rewritten as
4 + (−6x). If we were evaluating this expression, according to
x and then the 4 is added last. So to undo these operations
order of operations, -6 is rst multiplied to
and get
x
by itself, we work in the reverse order and rst undo the addition of 4 (by adding -4) and
then undo the multiplication by -6 (by multiplying by
4 − 6x
=
4 + (−6x)
=
−4
− 61 .)
8
8
−4
−6x = 4
1
1
− (−6x) = − (4)
6
6
4
x = −
6
2
x = −
3
2) Here we need to get rid of all the brackets and parentheses (by using the distributive property) so that
we can combine all the terms that have
x
in them.
CHAPTER 4.
12
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
x − 3[x − (2 − x)] = −2
x + (−3)[x + (−1)(2 + (−x))] = −2
Convert all the subtractions to adding by the
opposites.
x + (−3)[x + (−2) + x)] = −2
With multiple nested parentheses and brackets, we
work from the inside out.
x + (−3)[2x + (−2)] = −2
x + (−6x) + 6 = −2
x
Now that we only have one term with
can solve for
−5x + 6
−2
=
−6
x.
following the reverse of the order of operations, by
−6
adding -6 to both sides of the equation.
−5x = −8
= − 15 (−8)
− 15 (−5x)
Now we undo the multiplication by -5 by
multiplying by the reciprocal
x=
3) In order to get
x
in it, we
We rst undo the addition by 6,
− 15 .
8
5
by itself on one side of the equation, we need to combine all the
x's
this, we need to get rid of the parentheses as they prevent us from combining all the
1 + 4(x − 3) = 3(3 − 2x) − 2x
together. To do
x's.
To get rid of the parentheses, we distribute the 4
and 3 into
x−3
and
3 − 2x
respectively.
1 + 4x + (−12) = 9 + (−6)x + (−2)x
We need to get all the x's on one side of the
equation. Since the
4x + (−11) =
+8x
9 + (−8)x
+11
+11
+8x
−8x is added to 9, we undo
8x to both sides. This will
the addition by adding
remove the x term on the right side of the
equation. Likewise, we add 11 to get rid of the -11
on the left side of the equation.
12x = 20
1
= 12
(20)
5
x= 3
1
12 (12x)
4) Since we are solving for
x,
we treat the other variables,
must also be on the other side of the equation from
a(x − 2) = bx − 3
a(x + (−2)) = bx + (−3)
a
and
b,
like numbers, which means that they
x.
In order to combine all the x's, we need to get rid
of the parentheses by distributing in the
a
since
x
is inside the parentheses.
We get rid of the
−bx
ax + (−2)a =
−bx
+2a
bx + (−3)
−bx
+2a
ax − bx = −3 + 2a
bx,
on the right side, by adding
to both sides and we get rid of
left side, by adding
2a
(−2)a,
Now, although we cannot technically add
−bx,
on the
to both sides.
ax
and
recall that combining like terms was really
the distributive property used in the reverse
direction (see Sec.
??), which we call factoring.
Here we can factor out the
(a − b)x = −3 + 2a
1
− b)x = a−b
(−3 + 2a)
1
a−b (a
x.
−3 + 2a since
1
needs to be multiplied to the entire side, i.e.
a−b
to both terms.
We need the parentheses around
CHAPTER 4.
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
x=
4.4.2
13
−3+2a
a−b
Equations with Fractions
View the Video Tutorial for this section here.
Given an equation that contains fractions, the multiplication property of equality, Proposition 4.3.1, gives
us a technique for getting rid of the fractions. We will describe the technique through the following example.
First we nd the least common denominator (LCD)
of all the fractions on both sides of the equal sign,
x 5
3
− =
3 6
2
which in this case is 6
Now we multiply both sides of the equation by the
6
x 5
−
3 6
3
=6
2
LCD.
Remember that we must multiply the entire side by
x
5
=9
+6 −
6
3
6
6. That means that every term in the equation is
multiplied by 6.
Now that all the fractions are gone, we solve as we
have done before.
2x + (−5) =
+5
+5
2x = 14
= 12 (14)
x=7
1
2 (2x)
9
CHAPTER 4.
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
14
Summary of Technique:
When solving equations that contain fractions, we can use the multiplication property of equality to
remove all the fractions as follows:
•
Find the least common denominator (LCD) of all the fractions on both sides of the equation.
•
Multiply both sides of the equation by the LCD.
•
Note that this technique only works for equations, because the multiplication property of equality
only applied to equations.
Equation
Expression
We can apply the technique as shown
We cannot get rid of the fractions. Here,
above.
you make each fraction have the same
LCD and then carry out the subtraction.
3
x 5
− =
3 6
2
Example 4.4.2.
x 5
−
3 6
2 x
5
−
2 3
6
2x − 5
6
Solve.
2)
1)
5x x − 1
2x + 3
−
=
12
2
3
x
x−2
−5=
2
4
Solutions:
1)
Since this is an equation, we can get rid of the
fractions by multiplying through by the LCD, which
x
x−2
−5=
2
4
in this case is 4.
Remember that we must multiply the entire side of
4
4
x
x−2
−5 =4
2
4
x−2
2
+ 4(−5) = 4
the equation by 4.
x
4
2(x − 2) + (−20) = x
Since there are
x's
on both sides of the equation, we
need to distribute the 2 inside the parentheses so
that we can collect all the
x's
together on one side of
the equation.
2x + (−4) + (−20) = x
Since the
add
2x + (−24)
−2x
=
x
−2x
x's
−2x
2x
is added to -24, to undo the addition, we
to both sides. Note that this places all the
on the right side of the equation.
CHAPTER 4.
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
−24 = −x
−1(−24) = −1(−x)
24 = x
15
Recall that the multiplicative inverse of -1 is -1.
2)
Here the LCD is 12. But before we multiply by it,
keep in mind that the fraction bar implies
5x x − 1
2x + 3
−
=
12
2
3
parentheses around the numerator and denominator.
So when we multiply by 12, we must multiply the
entire numerator by 12.
12
5x x − 1
−
12
2
= 12
2x + 3
3
12
3 to
4, keep in mind that the 4 must still be multiplied to
When we reduce the right side of the equation
12
5x
12
+ (−12)
x−1
2
the whole numerator
= 4(2x + 3)
2x + 3.
Likewise, the subtraction on the left side of the
equation applies to the entire numerator, so it can be
viewed as
+(−1) (x−1)
2
which when multiplied by 12
gives us the -12.
5x + (−6)(x − 1) = 4(2x + 3)
5x + (−6x) + 6 = 8x + 12
−x + 6
−12
+x
=
8x + 12
−x
−12
−6 = 9x
1
1
9 (−6) = 9 (9x)
− 23 = x
4.4.3
Special Cases
4.4.3.1 Identity
In the above examples, when we solved the equations, we ended up with one answer. Now we will discuss a
couple of special situations that can arise while solving equations. We start by solving the following equation.
2x − 15
=
3(x − 5) − x
2x + (−15)
=
3x + (−15) + (−x)
2x + (−15)
=
−2x
−15
Note here that all the terms with
x
2x + (−15)
−2x
= −15
in them have disappeared. The question is then what is the solution to
the original equation. This is where we have to remember that the addition and multiplication property of
equality stated that when you add or multiply the same number to both sides of the equation, not only are
they still equal, but the solutions to the new equation and the old are the same. In other words, the values
x that make −15 = −15 are also the values of x that make 2x − 15 = 3(x − 5) − x. Since the equation
−15 = −15 is always true and does not depend on what value x is, this means that the original equation
will also always be true, no matter what x is. In other words, no matter what x value you choose, if you
of
work both the left and right side out separately, the numbers will always equal each other. In such cases, we
say that the solution to the equation
always true, an Identity.
2x − 15 = 3(x − 5) − x
is all real numbers. We call equations that are
CHAPTER 4.
Denition 4.4.1.
•
16
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
An equation that is always true is called an Identity.
An equation is an Identity when you try to solve it for the variable, but the
variable ends up being eliminated from the equation and you end up with the
same number on both sides of the equation, like
•
2 = 2, 0 = 0
or
−5 = −5.
For such equations, we say that the solution is all real numbers.
4.4.3.2 Inconsistent
The second special case occurs when the variable disappears from the equation, but what remains do not
equal each other. For example
2x − 5
=
3(x − 5) − x
2x + (−5)
=
3x + (−15) + (−x)
2x + (−5)
=
−5
=
−2x
2x + (−15)
−2x
−15
Since -5 never equals -15, this means that there is no
turn means that there is no
x
x
value that makes the equation
value that will make the equation
no matter what value you pick for
x,
2x − 5 = 3(x − 5) − x
−5 = −15,
which in
true. In other words,
if you work out the left and right side out separately, the two sides will
never give you the same number. In such cases, we say that there is no solution. We call equations that are
never true, inconsistent.
Denition 4.4.2.
•
An equation that is never true is called Inconsistent.
An equation is inconsistent when you try to solve it and the variable ends up
being eliminated from the equation and you end up with dierent numbers on
either side of the equation, like
•
0 = 1, −3 = 3,
or
2 = 5.
For such equations, we say that it has no solutions.
CHAPTER 4.
4.5
17
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Applications
4.5.1
Percentage
Recall that percentages are a way to talk about a part or piece of something. For example,
of something and
25%
50%
means half
means a quarter. Now, suppose someone came up to you and said if you gave him
your car, he would give you half. Would you do it? A natural question to ask would be half of what. Would
that not make a dierence in how you would answer the proposition? Suppose the person that came up to
you was Bill Gates and he was saying that he would give you half his wealth for your car. What if it was Bill
Gates saying that he would give you half his lunch? Notice that your answer will probably depend on what
he meant by half. If you are talking about a part or piece then you need to be thinking about or identifying
the whole from which you take the piece. So when we talk about percentages, the rst question that should
pop into your head is, percentage of what? In other words, what is the whole from which we are taking a
piece or part. The other thing to keep in mind with percentages is that the part or piece can be referred to
either as a percentage or the actual amount. For example, if you slice an apple pie into ten even slices, the
we can talk about
20%
of the pie (percentage of the whole that we want) or two slices (the actual amount of
the whole pie that we want.) Both are referring to the same amount of the pie. The formula for calculating
percentages is
Denition 4.5.1.
The formula for calculating percentages is
part is percentage of whole
part = percentage (%)
•
· whole
Remember that a percentage is not a number and so must be converted to its
decimal form before using it in the above formula.
•
If we know any two of the three quantities that make up the above formula (part,
percentage or whole), we can solve for the third quantity.
Example 4.5.1.
Write out the appropriate formula before solving the problem.
1) 12.5% of what is 7?
3) Suppose the total bill at a restaurant came to
$92.88 including an 8% sales tax.
2) If 62.5% of the students in the class are female
What the
price of the meal before tax?
and there are 12 males, how many females are
in the class?
Solutions:
1) Here 7 is the part and
.15 is the percentage and our unknown is the whole.
So if we let
x be the whole,
then
.125x = 7
1
1
(.125x) =
(7)
.125
.125
x = 56
2) Our problem here is that we are given the percentage of female students but the part given is the
number of male students. The 62.5% refers to the percentage of all students that are female. In other
words, the whole when talking about 62.5% is the total number of students which we do not know. So
in terms of females, all we know is the percentage. However, since the class can only be made up of
females and males, if 62.5% are female then
100 − 62.5
or 37.5% are males. Since we are told there are
12 male students, that represents the part when the percentage is 37.5%. Since we have the part and
the percentage, we can nd the whole. Note that whether we are talking about the 62.5% females or
the 37.5% males, the whole that those percentages refer to is the total number of students, both male
and female. So if we let
x
be the total number of students, then we have
12 =
12
=
.375
32 =
.375x
x
x
CHAPTER 4.
18
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
So there are 32 students in total. Since the problem asked for the number of females, we take the total
number of students and subtract the total number of males.
32 − 12 = 20
So there are 20 female students in the class.
3) In solving word problems, the most common mistake that students make is to try and write an equation
right away. The rst step should be to understand what the problem is talking about and to write
out in words the relationship between the various information given. For example, in this problem, we
are talking about the cost of a meal. Most of us understand that when you buy most items, you not
only pay the cost of the item, but you must also pay a sales tax on the item. So the total bill at the
restaurant is made up of the price of the meal plus the sales tax for that meal,
total cost
= cost
of the meal
+ sales
tax (8% of the meal cost)
We are given that the total cost of the meal is $92.88. If we let
92.88
=
x be the cost of the meal,
then we have
x + .08x
92.88 = 1.08x
92.88
= x
1.08
86 = x
So the cost of the meal before taxes is $86.00.
4.5.2
Mixture Problems
View the Video Tutorial for this section here.
To talk about mixture problems, we need to rst discuss what concentrations of various material (e.g.
sugar, salt or acid) in a medium (usually liquid) means. Suppose you go into a convenience store and buy a
20 oz fruit drink which claims that it is made with 5% real fruit juice (i.e. the drink has 5% concentration of
fruit juice.) If you drink the whole thing, then 5% of what you drank is fruit juice or another way to say it
is that you drank
.05(20) = 1
ounce of actual fruit juice. The other 19 ounces was water, sugar, and various
other chemicals. For your health, what is in the other 19 ounces would be important, but for our discussion
of solving mixture problems, what is important to remember is that we can talk about the fruit juice in the
fruit drink two dierent ways. Either as 5% of the whole drink or 1 ounce, the actual amount of the whole
drink that is fruit juice.
The mixture problems that we will deal with always involves mixing two solutions of dierent concentrations to get third, dierent concentration. The diagram used to organize the given information of each
problem will be the same. The dierence will be in the actual values used. So to be able to solve these types
of problems, one must rst understand how to build the diagram.
Example 4.5.2.
Be sure to write a sentence at the end to answer each question.
1) How many liters of a 15% acid solution should
2) How many gallons each of iced tea that contains
we mix to 8 liters of a 40% acid solution to ob-
35% sugar should be mixed with iced tea that
tain a 30% acid solution? Round the answer to
has 5% sugar to make 6 gallons of iced tea with
four decimal places.
10% sugar?
Solutions:
1) To solve mixture problems, we will always use the following diagram.
This row is just labeling each bottle.
bottle 1
bottle 2
bottle 3
15% acid
40% acid
30% acid
solution
solution
solution
Total amount of liquid in each bottle.
x
+
8
=
x+8
Actual amount of acid in each bottle.
.15x
+
.4(8)
=
.3(x + 8)
If we let
x
be the amount of the 15% acid solution mixed, then 15% of
x
or
.15x
is the actual
amount of acid in this bottle. So for each bottle, the actual amount of acid is found by multiplying
CHAPTER 4.
19
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
the corresponding percentage to the total amount of liquid in each bottle. Since we get the 30% acid
solution by mixing the other two, the total amount of liquid in the 30% bottle comes from the other
two bottles, i.e.
the total amount of liquid in the 30% bottle is the sum of the other two bottles.
Likewise, the acid in the 30% bottle can only come from the other two. This means that
.15x + .4(8)
=
.15x + 3.2
=
.8
.8
.15
5.3333
=
.15x
=
x
≈
x
−.15x
−2.4
.3(x + 8)
.3x + 2.4
−.15x
−2.4
You need to mix 5.3333 liters of the 15% acid solution to the 40% solution to get a 30% acid solution.
2) In this problem, we do not know how much of the two dierent iced teas we need to mix together.
However, we do know the total when mixed together must be 6 gallons, i.e. the sum of the two must
be 6. So if we let
x
6 − x must be the amount of
x + (6 − x) = 6 (Note that we could have let x be the
case 6 − x would have been the amount of the iced tea
be the amount of the iced tea with 35% sugar, then
the iced tea with 5% sugar that we mix in since
amount of the iced tea with 5% sugar, in which
with 35% sugar. You should try solving the problem this way to verify that you still get the same nal
answers. This is why it is important that you write a sentence at the end to state your answer as the
variable you solve for may or may not be what was asked for, depending on how you had set up the
problem.)
This row is just labeling each bottle.
Total amount of iced tea in each
bottle 1
bottle 2
35%
5%
bottle 3
10%
sugar
sugar
sugar
x
+
6−x
=
6
.35x
+
.05(6 − x)
=
.1(6)
bottle.
Actual amount of sugar in each
bottle of iced tea.
As with the previous example and all the mixture problems that we encounter, we get our equation from the actual amount of whatever concentration the percentages are based on (in this case
sugar.) Since the sugar in the nal mixture can only come from the two bottles that we mix together,
we have that
.35x + .05(6 − x)
= .1(6)
.35x + .3 − .05x = .6
.3x + .3
−.3
.3x
x
x
=
= .3
.3
=
.3
= 1
So we need to mix 1 gallon of iced tea with 35% sugar to
4.5.3
.6
−.3
6−1=5
gallons of iced tea with 5% sugar.
Distance/Rate Problems
View the Video Tutorial for this section here.
In order to solve what we call Distance/Rate problems, it is important to rst understand the relationship
between distance, rate, and time. Suppose a car is traveling at a constant speed (rate) 70 miles per hour
(mph). This means that every hour the car has gone another 70 miles in distance. So
CHAPTER 4.
20
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
number of hours traveling
distance traveled (in miles)
1
70
2
70 + 70 = 2(70)
3
z }| {
70 + 70 +70 = 3(70)
.
.
.
.
.
.
t
t(70)
2(70)
So, in this case, the distance traveled,
Denition 4.5.2.
d,
in time
t
hours is
d = 70t.
In general we say that
The formula relating distance, rate and time is
d=r·t
where
• r
is the constant rate (speed) at which the object travels.
• t
is the travel time, (the length of time the object has traveled at speed
• d
is the distance traveled in time
•
t
and at speed
r.)
r.
Note that the units for speed, time, and distance must be the same, e.g. if the
speed is in miles per hour, then time must be in hours and distance must be in
miles whereas if the speed was in feet per second, then time must be in seconds
and the distance must be in feet.
Just as with the mixture problems, the table we use to organize the information from any distance/rate
problem will be the same.
So understanding what the table says and how to build it, is the key step to
understanding how to solve these word problems.
Example 4.5.3.
Be sure to write a sentence at the end to answer the question.
will they meet?
1) Two hikers are on the opposite ends of a 15
mile trail that starts at the base and ends at
2) A passenger train leaves the station at noon
the top of a mountain. At 6:00 a.m. both hik-
and travels east at a rate of 50 m.p.h.
ers start their hike, with the hiker at the base
One
hour later, an express leaves the same station
of the mountain walking up the trail at 2 miles
chasing after the passenger train at 65 m.p.h.
per hour, while the hiker coming down from the
How long will it take the express to catch up to
top is walking down at 3 m.p.h. At what time
the passenger train?
Solutions:
1) For distance/rate problems, we build a table with one column being the speed, another column being
the time traveled, and a third column representing distance traveled. The distance column is always
lled by multiplying the rate (speed) and time.
rate
*
time
=
distance
rate
time
distance
in miles per hour
in hours
in miles
hiker at the base
2 m.p.h.
hiker at the top
3 m.p.h.
t
t
2t
3t
Since the hikers started and stopped at the same time, the time that they have each hiked is the
same. As we do not know that time, we let it be
t.
Recall that we said previously that the distance
in the above table is always lled by multiplying the rate and time. So
the hiker going to the base to the top of the mountain, while
3t
2t
is the distance traveled by
is the distance traveled by the hiker
coming down the mountain. The equation that we build to solve distance/rate problems will always
come from some relationship between the distances. In this case, since the total distance between the
two hikers is 15 miles, when they meet on the trail, the sum of the distances, traveled by each hiker,
should result in 15 miles. See gure below.
CHAPTER 4.
21
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
So, we have that
2t + 3t =
15
5t =
15
15
t =
5
t = 3
This means that it takes 3 hours for the two hikers to meet. Since they started at 6:00 a.m., they meet
at 6+3 or 9:00 a.m.
2) Since the express leaves one hour after the passenger train and they both stop when they meet, if we
let
t
be the total time that the passenger train has been traveling when the two meet, then the total
time the express train has been traveling must be one hour less, so
t − 1.
The table that we build for
this problem is as follows:
rate
*
time
=
distance
rate
time
distance
in miles per hour
in hours
in miles
passenger train
50 m.p.h.
express
65 m.p.h.
t
t−1
50t
65(t − 1)
Since both trains leave the same station and stop when they meet, the distance traveled by each
train is the same. So
50t =
50t =
−65t
65(t − 1)
65t − 65
−65t
−15t = −65
65
t =
15
13
t =
3
13
3 hours is the time the passenger train travels.
travels, we need to calculate
Note that
t =
t−1
=
=
=
So the express travels for
10
3 or
3 31
hours.
13
−1
3
13 3
−
3
3
10
3
To nd the total time the express
CHAPTER 4.
4.6
22
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Linear Inequalities in One Variable
The techniques used to solve linear inequalities are similar to those we used to solve equations. There are,
however, some subtleties in dealing with inequalities. One is that the solution will usually involve an innite
number of numbers and so we need a way to represent those answers. The other has to do with multiplying
by negative numbers.
4.6.1
Graphing Inequality Solutions on the Number Line
View the Video Tutorial for this section here.
We begin by looking at how to represent the solutions. Given two numbers, in comparing them, one of
the following three statements must be true.
•
The two numbers are equal (=) in value,
•
the rst number is greater than (>) the second number,
•
or the rst number is less than (<) the second number.
This is called the trichotomy of numbers. When we write
that is the only number that you can replace
x=2
x
x=2
it is understood that the solution is 2 as
with and still have a true statement.
We say that 2 is the solution because if you replace x with any other
number, the statement is not true.
(2) = 2
(3) = 2
(−2) = 2
Replace x with 2 and the statement is true.
Replace x with 3 and the statement is not true.
Replace x with -2 and the statement is not true.
However, if we have
x > 2, note
x>2
that will make the statement
that there are many numbers, in fact an innite quantity of numbers,
true. We call all these numbers the solutions to the inequality
x > 2.
When students are asked to think of all numbers greater than 2, they often think the answers must be whole
numbers,
3, 4, 5, . . .
However, the decimal number 2.01 is also greater than 2. Note that if you collect all the
real numbers between 2 and 3, there will be more there than all the whole numbers,
x>2
1, 2, 3, . . .
This means all numbers x such that they are greater than 2. There are an
innite quantity of numbers that will satisfy this statement.
(2.01) > 2
(2.001) > 2
(3) > 2
(2) > 2
Replace x with 2.01 and the statement is true.
Replace x with 2.001 and the statement is true
Replace x with 3 and the statement is true
Replace x with 2 and the statement is not true, since 2 is not greater than
2. This also shows us that although
x>2
has an innite number of
solutions, not every number will be a solution.
Since it is not possible to list all the numbers that satisfy the statement
x > 2,
another way to express
this statement is with a graph on the number line.
The hollow circle around the 2 indicates that we start at 2 but do not include it, but we do include all the
numbers that are close to 2 but greater than 2, such as 2.01, 2.001, etc.
and the line going to the right
indicates that we are talking about numbers greater than 2. Note that nowhere on the number line do we
write
x.
That is because it is understood that the whole graph is talking about
graphing the solutions on the number line is as follows.
x.
The general rule for
CHAPTER 4.
23
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Denition 4.6.1.
Inequality Signs (<, ≤, >, ≥)
When graphing the solution to inequalities on the number line, remember that on the
number line, anything to the left is smaller and anything to the right is bigger. In the
following, we assume that the variable is
• x<a
x
means all numbers less than
shade to the left of
and the number is
a,
so we use a hollow circle around
• x > a means all numbers greater than a,
shade to the right of a.
a
and shade to the right of
Example 4.6.1.
1)
and
so we use a lled circle around
so we use a hollow circle around
means all numbers greater than or equal to
around
a
a.
• x ≤ a means all numbers less than or equal to a,
a and shade to the left of a.
• x≥a
a.
a,
a and
so we use a lled circle
a.
Graph the solution on the number line.
x ≥ −3
2)
x<
5
3
3)
−3 ≥ x
Solutions:
1) Since we can include -3, we use a lled circle.
2) Note that there is no need to but tick marks at 0, 1, 2 and then place
5
to write is
3 and use a hollow circle.
5
3 between 1 and 2. All we need
3) It is usually easier to read the inequality with the variable on the left side.
expression around. Think of
−3 ≥ x
So we ip the entire
as being written on a transparent piece of paper and you ip the
paper to view the back side. You would see
x ≤ −3.
This is the same as
−3 ≥ x.
The way to check
that you ipped it correctly, is that whatever the inequality was pointing to, it should still point to it
after the ip. In this case, the pointy part of the inequality sign is going towards
x.
CHAPTER 4.
4.6.2
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
24
Addition/Multiplication Property of Inequalities
View the Video Tutorial for this section here.
Click here to watch a video on interval notation.
The addition property of inequality works just like the addition property of equality.
Proposition 4.6.1.
Addition Property of Inequality
If you add the same quantity to both sides of an inequality, the inequality is still true.
then
•
if a > b
a+c>b+c
then
if a < b
a+c<b+c
Note that the solutions to the new and old inequalities must also be the same.
Just like with equations, this makes sense if you think of a scale. Only this time, the scale is not balanced.
One side is heavier than the other. If you add or take away the same amount of weight from both sides of
the scale, the same side will still be heavier than the other.
The multiplication property of inequality is a little more subtle than its equality counterpart.
The
dierence occurs when dealing with multiply both sides by a negative number. For example,
2<5
is a true statement. If we multiply both sides by 3, we get
3(2) < 3(5)
6
< 15
which is also true. Now, if we had multiplied by -3 instead of 3, we get
−3(2) < −3(5)
−6
< −15
which is not true. To make the statement true, we must change the sign of the inequality to
−6 > −15
So, in general we have
Proposition 4.6.2.
Multiplication Property of Inequality
When multiplying both sides of an inequality by a nonzero number,
•
if the number you are multiplying is positive, then the inequality sign does not
change.
For
c>0
a<b
c·a<c·b
if
then
•
a>b
c·a>c·b
if
then
if the number you are multiplying is negative, then the inequality sign must
change (go from < to >or from > to <).
For
a<b
c·a>c·b
if
then
c<0
a>b
c·a<c·b
if
then
When solving inequalities, the general strategy is the same as for equations. We must get the variable
that we want to solve for by itself on one side of the inequality sign.
Example 4.6.2.
Solve. Graph your solution on the number line.
CHAPTER 4.
25
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
1)
2)
3)
x−2
2
1
> x−
6
3
2
1 − 4x ≤ 5
3x + 5 < x − 3
Solutions:
1) As with equations, we must get all the terms with variables on one side of the inequality.
right side
left side
z }| {
3x + 5 <
z }| {
x−3
To get all the
x's
to the left side, we add
−x
to both
sides. To also add -5 to both sides so that only the
3x + 5
−x
−5
<
x + (−3)
−x
−5
2x < −8
x's
remain on the left side.
Note that since we added (the
−x
and -5), the
inequality sign does not change.
1
2 (2x)
< 12 (−8)
x < −4
So the graph of the solution on the number line is
2)
1 − 4x ≤ 5
1 + (−4)x
−1
≤
5
−1
−4x ≤ 4
Since we added the -1 to both sides, the inequality
sign does not change.
− 14 (−4x)
≥
− 14 (4)
Since we multiplied both sides by a negative number,
− 14 ,
the inequality sign has changed direction.
x ≥ −1
So the graph of the solution on the number line is
3) As with the equations, we can use the multiplication property of inequalities to get rid of the fractions
by multiplying both sides by the least common denominator (LCD).
CHAPTER 4.
26
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
Multiply both sides by the LCD, which is 6 in this
6
x−2
6
>6
2
1
x−
3
2
case. Since we are multiplying by a positive number,
the inequality sign does not change directions.
Remember that we must distribute the 6 in on the
x−2>6
1
2
3 x and 2 are two separate terms,
x−2
where as the
6 is one term (fraction) and so we
can cancel the 6's out right away.
right side because
1
2
x + (−6)
3
2
x − 2 > 2(2x) + (−3)
x + (−2) >
−4x
4x + (−3)
−4x
+2
+2
−3x > −1
< − 13 (−1)
− 13 (−3x)
x<
Since we multiply both sides by a negative number,
− 13 ,
the inequality sign has changed direction.
1
3
So the graph of the solution on the number line is
4.6.3
Compound Inequalities
To talk about compound inequalities, we must rst talk about the conjunctions and and or. These words
have mathematical meaning when we are talking about a collection of things, such as numbers. For example,
x>3
we are talking about all numbers
and
x<7
x that are greater than 3 and at the same time less than 7.
In other words,
we are talking about all the numbers between 3 and 7. We refer to the word and as intersection. Why we
call it intersection can been seen by graphing the above two inequalities on the number line.
The line on top represents all numbers less than 7 while the line on bottom represents all numbers greater
than 3. So all the numbers that are both greater than 3 and less than 7 are where the two lines overlap
(intersect) each other, the shaded region on the graph. So the nal graph that represents
x>3
and
x<7
is
This can also be written with the double inequality sign
3<x<7
The word or is referred to as a union.
x < −3
or
x>5
means all number that are either less than -3 or greater than 5. This includes all the numbers described
by both inequalities, hence the union of both answers.
inequalities.
The graph would simple be the graph of both
CHAPTER 4.
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
27
So we can summarize the above as follows.
Denition 4.6.2.
Intersection (and)/Union (or)
and (Intersection)
•
means that both conditions connected by the word and must be satised.
•
it is referred to as an intersection and is denoted by the symbol
•
the combined notation
T
.
a<x<b
means
x>a
and
x < b.
Note that if
a
is not smaller than
b,
the above makes no
sense.
or (Union)
•
means that either conditions connected by the word or can be satised.
•
it is referred to as a union, and is denoted by the symbol
•
unlike the intersection (and), there is no combined notation for unions.
Example 4.6.3.
2)
S
x > −3 x < 5
T
x < 2 x > 52
3)
2x − 5 ≤ 3x + 2
1)
S
.
Solve and graph your solutions on the number line.
T
2(x + 6) + 1 < 3 − x
4)
−2 < 3 − 2x ≤ 7
5)
3x ≤ 5x + 3
S
18 ≤ 4 − 3(x − 3)
Solutions:
1) Remember that union,
S
, means or. So if we graph both inequalities
we see that the two lines cover the entire number line. This means that the answer is all real numbers.
Graphically, we shade the entire number line for our answer.
x is already by itself on one side of the inequality, we
= 2 21 > 2, which means that 52 is to the right of 2 on
2) As with the previous problem, since the variable
simply graph both of them. Remember that
5
2
the number line.
Since
T
mean intersection and we can see that the two lines to not intersect anywhere, this means
5
2 is bigger than 2, there
5
is no number that is bigger than
2 and at the same time smaller than 2. So there is no graph for this
one. We simply say that there is no solution or we can use the symbol ∅, which is called the empty set
that there is no solution to the inequalities. Note that this makes sense, since
(i.e. the collection of all the solutions to this problem is empty.)
CHAPTER 4.
28
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
3) Here we need to solve each inequality for
x
rst.
2x − 5 ≤ 3x + 2
T
2(x + 6) + 1 < 3 − x
−3x + 5 − 3x + 5
T
2x + 12 + 1 < 3 − x
−x ≥ 7
T
2x + 13 < 3 + (−x)
+x
−13
−13
+x
−1(−x) ≥ −1(7)
x ≥ −7
T
x ≥ −7
T
Now we graph the two inequalities. Note that
3x < −10
1
1
(3x) < (−10)
3
3
10
x<−
3
1
− 10
3 = −3 3 > −7.
So
− 10
3
is to the right of -7 on the
number line.
The answer is then all the numbers between -7 and
− 10
3 ,
including -7 but not including
− 10
3 .
4) Recall that the double inequality sign means intersection (and). The goal here is to get the
in the middle.
left side
z}|{
−2
middle
right side
z }| {
< 3 − 2x ≤
z}|{
7
x
by itself
Note that there are three sides to this problem. The
addition and multiplication properties still apply,
only we apply them to all three sides instead of two.
To undo the 3 we add -3 to all three sides.
−2 <
3 − 2x ≤ 7
−3
−3
−3
−5 < −2x ≤ 4
− 12 (−5) > − 12 (−2x) ≥ − 12 (4)
To undo the multiplication by -2, we multiply by
to all three sides. Since we are multiplying by a
negative number, the equality signs must change
direction.
5
2
> x ≥ −2
−2 ≤ x < 52
We ip the entire inequality around.
So the answers are all numbers greater than or equal to -2 and less than
5
2.
− 12
CHAPTER 4.
LINEAR EQUATIONS AND INEQUALITIES IN ONE VARIABLE
5) We solve the two inequalities for
29
x.
3x ≤ 5x + 3
S
−2x ≤ 3
1
1
− (−2x) ≥ − (3)
2
2
3
x≥−
2
S
−5x
−5x
18 ≤ 4 − 3(x − 3)
18 ≤ 4 + (−3)(x + (−3))
x≥−
Since
− 32 > − 53 , − 32
number line is
is to the right of
− 35
3
2
S
18 ≤ 4 + (−3x) + 9
18 ≤ 13 + (−3x)
−13
S
−13
5 ≤ −3x
1
1
− (5) ≥ − (−3x)
3
3
5
− ≥x
3
5
x≤−
3
on the number line. So the graph of the solution on the