Linear Equations and Word Problems
Algebraic Expressions
We will define the term variable which is analogous to a pronoun in grammar.
Definition
A variable
is a symbol that represents a number
that may change depending on the circumstance. 
Examples
We often use the variable t to represent time. We can
use the variable h to represent height. We use the letter
x for the
variable when the variable stands for a general number without context.
Definition
A constant
is a symbol that represents a known fixed
number. 
Examples
c = 3 x 10^{8
} p
= 3.14...
Definition
An algebraic expression
is an expression
consisting of constants, variables, and symbols of operators. 
Examples

m c^{2
}

p r^{2
}
3x^{2} +2x  1
Definition
A term
is an expression that is related by addition or subtraction. 
Example
In
3x^{2} +2x 
1
3x^{2} ^{
}2x and
1
are the three terms. 3, 2,
1 are called
the coefficients of
x^{2}, x, and
1 respectively. We say two
terms are like terms if they differ only in their coefficient.
Example
3x^{2 }
and
5x^{2}
are like terms. We can add like terms by adding their coefficients.
3x^{2 }+ 5x^{2}
= 8x^{2}
Exercise
Simplify the expression
5x^{2}
 2p + 3x^{2} + 5s^{2} + 2p 
3(2x^{2} 3s^{2})
Linear Equations
Recall that an expression is a statement involving x's, constants, and operators.
An expression is like the subject of a sentence. An equation
is more like a complete sentence which as a subject, verb and predicate.
The verb will typically be an "= " sign.
Example
3x + 1 = 4
is an example of an equation.
When we have an equation, we can attempt to solve it that is to find which x values
satisfy the equation. An x value satisfies an equation if when we
substitute the value in the equation for x, the equation is true. To solve for the variable
x means that we have an equation of the form.
x = a number.
The set of all x that satisfy the equation is called the solution
set of the equation.
To solve we isolate x by

Use the distributive property.

Combine like terms.
Subtract all the x terms from the right hand side "Do
unto the left as you would do unto the right."

Subtract all the constant terms from the left hand side.

Divide by the coefficient in front of x.
Example:
Solve the following equation for x.
3(2x  1) + 4x =
2(4x + 2)
Solution:

6x  3 + 4x = 8x + 4

10x  3 = 8x + 4

2x  3 = 4

2x = 7

x = 7/2
In the above example
we say that
x = 7/2
is a solution to the equation.
Example
Solve
3(2x  1)  2(3x + 1) = 0
Solution
6x  3  6x  2
= 0 (Distributed the
3 and the 2)
0  5 = 0
5 = 0
Since this can never happen we say that the equation is a contradiction.
Example:
Solve
2(x  3)  2x + 8 = 2
Solution:
2x  6  2x + 8
= 2 (Distributed the
2)
0 + 2 = 2
(Combined the
2x  2x and the
6 + 8)
2 =
2
Since this is always true, we say that the equation is an identity.
Problem Solving With Formulas
A formula is an equation that relates real world quantities.
Examples
P =
2l + 2w
is the formula for the perimeter P of a rectangle given the length
l and width w.
d =
rt
is the formula for the distance traveled d given the speed
s
and the time t.
A =
P + Prt
is the formula for the amount A in a bank account
t years after
P dollars is put in at an interest rate of r.
V = pr^{2}h
is the formula for the volume V of a cylinder of radius
r and height h, where p @ 3.14.
h
A =
(b_{1} + b_{2})
2
is the formula for the area A of
a trapazoid with height h and bases b_{1} and
b_{2}.
We say that a formula is solved for a variable x if the equation becomes
x =
stuff
where the left hand side does not include any x's.
Example
Solve
C = 2pr
for r then determine the radius of a circle with circumference
4.
Solution
Divide both sides by 2p:
C
= r
2p
Use the reflexive property to get
C
r =
2p
Now plug in 4 for C to obtain
4
2
r =
=
2p
p
Steps for Solving a Word Problem

Read the problem, sketch the proper picture, and label variables.

Write down what the answer should look like.

Come up with the appropriate formula.

Solve for the needed variable.

Plug in the known numbers.

Answer the question.

Example:
A pile of sand has the shape of a right circular cone.
Find the height of the pile if it contains 100 cc of sand and the radius
is 5 cm.
Solution:


The height of the pile is ________ cm.

We use the formula for the volume of a right circular cone:
V =
1/3 pr^{2}h.

Multiply by 3 on both sides to get
3V =
pr^{2}h.
Divide both sides by pr^{2} to obtain
3V
3V
= h or
h =
pr^{2}
pr^{2}

3(100)
12
h =
=
p5^{2}
p

The height of the pile is 12/p cm.
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