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 108

p  =  3.14...

 Definition   An  algebraic expression is an expression consisting of constants, variables, and symbols of operators.

Examples

1. m c2

2. p r2

3. 3x2 +2x - 1

 Definition A term is an expression that is related by addition or subtraction.

Example

In

3x2 +2x - 1

3x2      2x      and      1

are the three terms.  3, 2, -1 are called the coefficients of x2, x, and 1 respectively. We say two terms are like terms if they differ only in their coefficient.

Example

3x2 and 5x2

are like terms.  We can add like terms by adding their coefficients.

3x2 + 5x2  =  8x2

Exercise

Simplify the expression

5x2 - 2p + 3x2  + 5s2 + 2p - 3(2x2 -3s2)

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

1. Use the distributive property.

2. Combine like terms.

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

4. Subtract all the constant terms from the left hand side.

5. Divide by the coefficient in front of x.

Example:

Solve the following equation for x.

3(2x - 1)  + 4x   =   2(4x + 2)

Solution:

1. 6x - 3 + 4x   =  8x + 4

2. 10x - 3  =  8x + 4

3. 2x - 3  =  4

4. 2x  =  7

5. x  =  7/2

 Definition:   An equation that is true for some x but not for others is called a conditional equation.  The value of x that makes the equation true is called the solution of the equation.

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  =  pr2

is the formula for the volume V of a cylinder of radius r and height h, where p @ 3.14.

h
A  =         (b1 + b2
2

is the formula for the area A of a trapazoid with height h and bases b1 and b2.

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:

1. The height of the pile is ________ cm.

2. We use the formula for the volume of a right circular cone:

V  =  1/3 pr2h.

3. Multiply by 3 on both sides to get

3V  =  pr2h.

Divide both sides by pr2 to obtain

3V                                   3V
=  h
or      h  =
pr2                                   pr2

4.             3(100)              12
h =                       =
p52                 p

5. The height of the pile is 12/p cm.