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```College Algebra
Sixth Edition
James Stewart  Lothar Redlin

Saleem Watson
P Prerequisites
P.8
Solving Basic Equations
Overview
Equations are the basic mathematical tool for
solving real-world problems.
In this section, we learn
• How to solve equations.
Equations
An equation is a statement that two
mathematical expressions are equal.
• For example:
3+5=8
Variables
Most equations that we study in algebra
contain variables, which are symbols
(usually letters) that stand for numbers.
• In the equation 4x + 7 = 19, the letter x
is the variable.
• We think of x as the “unknown” in the equation.
• Our goal is to find the value of x that makes
the equation true.
Solving the Equation
The values of the unknown that make
the equation true are called the solutions
or roots of the equation.
• The process of finding the solutions is called
solving the equation.
Equivalent Equations
Two equations with exactly the same
solutions are called equivalent equations.
• To solve an equation, we try to find a simpler,
equivalent equation in which the variable stands
alone on one side of the equal sign.
Properties of Equality
We use the following properties to solve
an equation.
• A, B, and C stand for any algebraic expressions.
• The symbol  means “is equivalent to.”
Properties of Equality
These properties require that you perform
the same operation on both sides of an
equation when solving it.
• Thus if we say “add 4” when solving an equation,
that is just a short way of saying “add 4 to each
side of the equation.”
Solving Linear Equations
Linear Equation
The simplest type of equation is a linear
equation, or first-degree equation.
• It is an equation in which each term is either
a constant or a nonzero multiple of the variable.
Linear Equation—Definition
A linear equation in one variable is
an equation equivalent to one of the form
ax + b = 0
where:
• a and b are real numbers.
• x is the variable.
Linear Equations
These examples illustrate the difference
between linear and nonlinear equations.
Linear
Equations
Nonlinear
Equations
4x – 5 = 3
x2 + 2x = 8 Contains the square
of the variable
2x = ½x – 7
Reason for
Being Nonlinear
x  6x  0 Contains the square root
of the variable
x – 6 = x/3
3/x – 2x = 1 Contains the reciprocal
of the variable
E.g. 1—Solving a Linear Equation
Solve the equation
7x – 4 = 3x + 8
• We solve this by changing it to an equivalent
equation with all terms that have the variable x
on one side and all constant terms on the other.
E.g. 1—Solving a Linear Equation
7x – 4 = 3x + 8
(Given equation)
(7x – 4) + 4 = (3x + 8) + 4
7x = 3x + 12
(Simplify)
7x – 3x = (3x + 12) – 3x
(Subtract 3x)
4x = 12
(Simplify)
¼ . 4x = ¼ . 12
(Multiply by ¼)
x=3
(Simplify)
We do so in many examples.
• In these checks, LHS stands for “left-hand side”
and RHS stands for “right-hand side” of the original
equation.
We check the answer of Example 1.
x = 3:
• LHS = 7(3) – 4 = 17
• RHS = 3(3) + 8 = 17
• LHS = RHS
E.g. 2—Solving an Equation That Involves Fractions
Solve the equation
x 2 3
  x
6 3 4
• The LCD of the denominators 6, 3, and 4 is 12.
• So, we first multiply each side of the equation by
12 to clear denominators.
E.g. 2—Solving an Equation Involving Fractions
x 2
3

12     12
x
4
6 3
2x  8  9 x
8  7x
8
x
7
An Equation that Simplifies to a Linear Equation
In the next example, we solve an equation
that doesn’t look like a linear equation, but it
simplifies to one when we multiply by the
LCD.
E.g. 3—An Equation Involving Fractional Expressions
Solve the equation
1
1
x 3

 2
x 1 x  2 x  x  2
• The LCD of the fractional expressions is
(x + 1)(x – 2) = x2 – x – 2.
• So, as long as x ≠ –1 and x ≠ 2, we can multiply
both sides by the LCD.
E.g. 3—Equation Involving Fractional Expressions
1
1 
x 3 


( x  1)( x  2) 

  ( x  1)( x  2)  2

 x 1 x  2 
 x  x 2
( x  2)  ( x  1)  x  3
2x  1  x  3
x4
Extraneous Solutions
• Even if you never make a mistake in your
calculations.
• This is because you sometimes end up with
extraneous solutions.
Extraneous Solutions
Extraneous solutions are potential solutions
that do not satisfy the original equation.
• The next example shows how this can happen.
E.g. 4—An Equation with No Solution
Solve the equation
5
x 1
2

x4 x4
• First, we multiply each side by the common
denominator, which is x – 4.
E.g. 4—An Equation with No Solution
5 
x 1


( x  4)  2 
  ( x  4) 

x 4

 x 4
2( x  4)  5  x  1
2x  8  5  x  1
2x  3  x  1
2x  x  4
x4
E.g. 4—An Equation with No Solution
Now, we try to substitute x = 4 back
into the original equation.
• We would be dividing by 0, which is impossible.
• So this equation has no solution.
Extraneous Solutions
The first step in the preceding solution,
multiplying by x – 4, had the effect of
multiplying by 0.
• Do you see why?
• Multiplying each side of an equation by an
expression that contains the variable may
introduce extraneous solutions.
• That is why it is important to check every answer.
Solving Power Equations
Power Equations
Linear equations have variables only to the
first power.
Now let’s consider some equations that
involve squares, cubes, and other powers
of the variable.
• Such equations will be studied more extensively in
Sections 1.6 and 1.7.
Power Equation
Here we just consider basic equations
that can be simplified into the form
Xn = a
• Equations of this form are called power
equations
• They are solved by taking radicals of both
sides of the equation.
Solving a Power Equation
The power equation Xn = a has the
solution
X= na
if n is odd
X = n a
if n is even and a ≥ 0
If n is even and a < 0, the equation has
no real solution.
Examples of Solving a Power Equation
Here are some examples of solving
power equations.
• The equation x5 = 32 has only one real
solution: x = 5 32 = 2.
• The equation x4 = 16 has two real solutions:
x =  4 16 = ±2.
Examples of Solving a Power Equation
Here are some examples of solving
power equations.
• The equation x5 = –32 has only one real
solution: x = 5 32 = –2.
• The equation x4 = –16 has no real solutions
4
because 16 does not exist.
E.g. 5—Solving Power Equations
Solve each equation.
(a) x2 – 5 = 0
(b) (x – 4)2 = 5
E.g. 5—Solving Power Equations
Example (a)
x2 – 5 = 0
x2 = 5
x=± 5
The solutions are x = 5 and x = – 5 .
E.g. 5—Solving Power Equations
Example (b)
We can take the square root of each side
of this equation as well.
(x – 4)2 = 5
x–4=± 5
x=4± 5
(Take the square root)
The solutions are x = 4 + 5 and x = 4 – 5 .
E.g. 6—Solving Power Equations
Find all real solutions for each equation.
(a) x3 = – 8
(b) 16x4 = 81
E.g. 6—Solving Power Equations
Example (a)
Since every real number has exactly
one real cube root, we can solve this
equation by taking the cube root of
each side.
(x3)1/3 = (–8)1/3
x = –2
E.g. 6—Solving Power Equations
Example (b)
Here we must remember that if n is even,
then every positive real number has two real
nth roots.
• A positive one and a negative one.
• If n is even, the equation xn = c (c > 0) has two
solutions, x = c1/n and x = –c1/n.
E.g. 6—Solving Power Equations
81
x 
16
4
1/ 4
81 

(x )    
 16 
3
x
2
4 1/ 4
Example (b)
Equations with Fractional Power
The next example shows how to solve an
equation that involves a fractional power of
the variable.
E.g. 7— Solving an Equation with a Fractional Power
Solve the equation
5x2/3 – 2 = 43
• The idea is to first isolate the term with the
fractional exponent, then raise both sides of the
equation to the reciprocal of that exponent.
• If n is even, the equation xn/m = c has two solutions,
x = cm/n and x = –cm/n.
E.g. 7— Solving an Equation with a Fractional Power
5x2/3 – 2 = 43
5x2/3 = 45
x2/3 = 9
x = ±93/2
x = ±27
The solutions are x = 27 and x = –27.
Solving for One Variable in
Terms of Others
Several Variables
Many formulas in the sciences involve
several variables.
It is often necessary to express one
of the variables in terms of the others.
• In the next example, we solve for a variable
in Newton’s Law of Gravity.
E.g. 8—Solving for One Variable in Terms of Others
Solve for the variable M in
mM
F G 2
r
• The equation involves more than one variable.
• Still, we solve it as usual by isolating M on
one side and treating the other variables as
we would numbers.
E.g. 8—Solving for One Variable in Terms of Others
Gm 

F   2 M
(Factor M from RHS)
 r 
2
2
r
r



  Gm 
 Gm  F   Gm   r 2  M (Multiply by reciprocal of





2
r F
M
Gm
r 2F
The solution is M 
.
Gm
( Simplify)
Gm
r
2
)
E.g. 9—Solving for One Variable in Terms of Others
The surface area A of the closed rectangular
box shown can be calculated from the length
l, the width w, and the height h according to
the formula:
A = 2lw + 2wh + 2lh
• Solve for w in terms
of the other variables
in the equation.
E.g. 9—Solving for One Variable in Terms of Others
The equation involves more than one
variable.
Still, we solve it as usual by isolating w
on one side, treating the other variables
as we would numbers.
E.g. 9—Solving for One Variable in Terms of Others
A = (2lw + 2wh) + 2lh
(Collect terms
involving w)
A – 2lh = 2lw + 2wh
(Subtract 2lh)
A – 2lh = (2l + 2h)w
(Factor w from RHS)
A  2l h
w
2l  2h
(Divide by 2l + 2h)
A  2l h
The solution is w 
2l  2h
```
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