In an equation such as  a = b / c  being able to solve for a, b, or c correctly is extremely important.  The most common mistake in this situation is thinking that the solution in the case of solving for c is not recognizing that  c = b / a .  Way too often students think that the value for c is equal to a / b.  The most basic rule for solving algebraic equations of any size or complexity is that:

"You must always do the same thing to both sides of an equation, regardless of what operation you are applying."

When a student thinks that c = a / b in the above equation rather than c = b / a, it is because they did NOT apply this rule.

To begin consider the ratio and proportion equation a / b = c / d. 

Step 1. When solving for any variable in this type of equation first cross multiply.

This means that a is multiplied times d, and b is multiplied times c.

This yields a d = b c

Step 2. To solve for any one of the four variables, the variable being solved for must be isolated by the use of division.

For Example, if the variable being solved for was d, both sides of the equation would be divided by a.

This yields d = b c / a
 

As a similar example, consider the following.

The equation  a = b / c  can be solved for c by the following two steps. 

Step 1.  Multiply both sides of the equation by c

This yields  c a = c b / c

This simplifies to  c a = b

Step 2.  Divide both sides of this equation by c

This yields  c a / c  = b / c

This simplifies to  a = b / c

Thus solving for c  in the expression (equation)  a = b / c

results in the expression (equation)  c = b / a

[Note: For convenience a could be written over 1, as in a / 1 = b / c, and c could be solved for using a ratio and proportion approach.]
 

Additional examples of solving for variables:

As stated above the most important rule to focus on in solving any equation is to avoid destroying an equality by doing some operation to one side that you do not do to the other side.

To see the significance of this statement look at the following examples.

Example 1: solve for "v" in the equation v = d / t

 The "d" is a numerator (a value in the top of a fraction)
 The "t" is a denominator (a value in the bottom of a fraction)
 The "v" is equal to "d" divided by "t"

Example 2: solve for "d" in the equation v = d / t

 The "d" is a numerator
 The "t" is a denominator
 "v" is equal to "d" divided by "t"

 "d" can be solved for by multiplying both sides by "t"
 This gives us v t = d or d = v t

Example 3: solve for t in the equation v = d / t

 The "d" is a numerator
 The "t" is a denominator
 The "v" is equal to "d" divided by "t"

 "t" can be solved for by
  1st multiplying both sides by "t" which gives
   v t = d
  2nd dividing both sides by "v" which gives
   t = d / v

The second rule is to be patient when solving equations which will require that you use several steps.

Example 4: Solve for "t" in the equation v(f) = v(i) + at

 There are no fractions here but there is an addition
 of unlike variables on the right side of the equation.

 "t" can be solved for by
  1st subtracting "v(i)" from both sides which gives
   v(f) - v(i) = at
  2nd divide both sides by "a" which gives
   a = v(f) - v(i) / a = t

This form of notation is a way of writing numbers using exponents to show how large or small a number is in terms of powers of ten.

Powers of ten are a means of writing what we might think of as everyday numbers as powers of ten such as 234 = 2.34 x 10^2, which can be read as two point three four times ten raised to the second power or ten squared.

The powers of ten will be written using the "carrot" symbol to indicate the exponent of the power of ten, as in one time ten to the third power is written as 1 x 10^3.

Example 1: Simple powers of ten

1 x 10^9 = 1,000,000,000
1 x 10^6 = 1,000,000
1 x 10^3 = 1,000
1 x 10^1 = 10
1 x 10^0 = 1
1 x 10^-1 = 1 / 10
1 x 10^-3 = 1 / 1000
1 x 10^-6 = 1 / 1,000,000
1 x 10^-9 = 1 / 1,000,000,000

Example 2: Converting numbers into scientific notation

The numbers must be written according to a simple rule. The none power of ten portion of the number may only have
one digit before the decimal. i.e. 22.65 x 10^3 would be incorrect while 2.265 x 10^4 is correct

314 = 3.14 x 10^2

7,183 = 7.183 x 10^3

230,100 = 2.301 x 10^5

0.0204 = 2.04 x 10^-2

0.0004607 = 4.067 x 10^-4

0.00000198 = 1.98 x 10^-6

Example 3: Converting from scientific notation

6.9702 x 10^5 = 69,702

4.599 x 10^3 = 4,599

2.0023 x 10^1 = 20.023

4.205 x 10^-2 = 0.04205

7.45 x 10^-4 = 0.000745

3.97 x 10^-5 = 0.0000397

1-3 Adding and Subtracting in Scientific Notation

Though calculators can do this for us, it is always important to see the math process behind the operation.

The fundamental rule when adding and subtracting numbers in scientific notation (exponential notation) is to be sure that the powers of ten for each number being added (or subtracted) is the same.

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Adding numbers in scientific notation

 Add the two numbers 4.55 x 10^2 plus 3.77 x 10^3

 1st change one of the values so that it has the same power of ten

 i.e. 0.455 x 10^3 plus 3.77 x 10^3

 2nd add the none exponent values 0.455 + 3.77 = 4.225

 3rd write the sum followed by the power of ten

 Answer: 4.225 x 10^3

Example 2: Subtracting numbers in scientific notation

 Subtract the two numbers 7.65 x 10^4 minus 4.32 x 10^3

 1st change one of the values so that it has the same power of ten

 i.e. 7.65 x 10^4 minus 0.432 x 10^4

 2nd subtract the none exponent values 7.65 - 0.432 = 7.218

 3rd write the difference followed by the power of ten

 Answer: 7.218 x 10^4

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Multiplying and Dividing in Scientific Notation

Though calculators can do this for us, it is always important to see the math process behind the operation.

The process of multiplying and dividing numbers written in scientific notation involves focusing on the exponents themselves. In the process of multiplication the exponents are added while in the process of division the exponents are
subtracted.

Example 1: Multiplying numbers in scientific notation

 Multiply the two numbers 3.0 x 10^4 times 6.0 x 10^5

 1st multiply the none powers of ten 3.0 x 6.0 = 18

 2nd add the exponents of the powers of ten 4 + 5 = 9

 3rd recombine the information as a single answer

 Answer: 18 x 10^9 which is correctly written as 1.8 x 10^10

Example 2: Dividing numbers in scientific notation

 Divide the two numbers 8.0 x 10^6 divided by 2.0 x 10^3

 1st divide the none powers of ten 8.0 / 2.0 = 4

 2nd subtract the exponents of the powers of ten 6 - 3 = 3

 3rd recombine the information as a single answer

 Answer: 4 x 10^3 which is correctly written

The metric system was developed around 1775 by French scientists. It is convenient to use because its units are related by powers of ten. A standardized system now exists worldwide. It is referred to as the International System (SI) of weights and measures.

Traditionally in the study of physics, two systems of measure have been used. They are referred to as the MKS and the CGS systems.

MKS system meter - kilogram - second

CGS system centimeter - gram - second

Many textbooks emphasize the MKS system

The Metric System: Four basic units of measure are displayed.

Length:  base unit = meter (fundamental unit)*
Mass:  base unit = gram (fundamental unit)*
Time:  base unit = second (fundamental unit)*
Volume:  base unit = liter (derived unit)*

* Units of measure are divided into two categories called the Fundamental unit and the Derived Unit.

Fundamental Unit: These are units of measure obtained directly from measurement observations using SI units and cannot be obtained indirectly from any simpler or more basic units of measure. Mass, length, and time are examples of fundamental units.

Derived Unit: These are units of measure which may be able to be observed through direct measurement (i.e. volume can be observed directly by visibly observing the reading off of the side of a graduate cylinder.), but also can be obtained by derivation using simpler fundamental units of measure. As an example consider that volume can be obtained from length by the formula V = L x W x H.

Example: Length of course is a length measurement (L), but so are width (W) and height (H) length measurements simply oriented in different directions. And it follows then that volume, which is a multiplication product of the three, is derived from length.

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Metric Prefixes - Units of Measure

Prefix Symbol Fractional Equivalent Example (using the meter)

Kilo K 1000 x base unit kilometer (Km)

hecto h 100 x base unit  hectometer (hm)

deka da 10 x base unit  dekameter (dam)

___ ___ 1 x base unit  meter (m)

deci d 1/10 x base unit decimeter (dm)

centi c 1/100 x base unit centimeter (cm)

milli m 1/1000 x base unit millimeter (mm)

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Additional Units:

Prefix (Symbol) Fractional Equivalent Example (using the meter)

giga (G) 1 x 10^9 x base unit gigameter (Gm)

mega (M) 1 x 10^6 x base unit megameter (Mm)

micro (m *) 1/1 x 10^-6 x base unit micrometer (mm *)

nano (n) 1/1 x 10^-9 x base unit nanometer (nm)

pico (p) 1/1 x 10^-12 x base unit picometer (pm)

* This is the Greek letter mu and not the English letter u.

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Common Metric - English Equivalents:

1 m = 39.37 in 1 cm = 0.394 in 1 in = 2.54 cm

1 kg = 2.21 lb 1 lb = 454 g 1 L = 1.057 qt

1 oz = 28.4 g 1 in^3 = 16.5 cm^3 1 gal = 3.78 L

1-8 Metric Conversion - Metric Sliding Scale Model

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Conversions between metric units:

The metric system is very easily manipulated as can be seen by the following two examples. Converting between two units of the same category of measurement can be done simply by moving the position of the decimal place at the same time the prefix is changed.  As the unit of measure is increased the decimal is moved to the left making the numerical value smaller proportionally. As the unit of measure is decreased the decimal is moved to the right making the numerical value larger proportionally. Look at the following two examples of this concept.

Example 1:  Consider the measurement 2.5 m and the variations of the same length measurement expressed in other metric units of length.

km                 hm              dam          b. u.*       dm            cm             mm

0.0025          0.025          0.25          2.5          25.0          250.          2,500.

*b. u. is the base unit of the category of measurement.  In this case the unit is the meter.

Example 2:  Consider the measurement 4,200 m and the variations of the same length measurement expressed in other metric units of length.

km          hm           dam           b. u.            dm                 cm                    mm

4.2          42.          420.          4,200.          42,000.          420,000.          4,200,000.

Student who study doing conversions between units within a measurement system such as Meteric to Metric or between measuring systems such as the Metric to English and English to Metric conversions usually can handle conversion factors that are one dimensional.  However, a problem often arises when conversion factors are needed for two and three dimensional problems.  The following information compares handling units in one, two, and three dimensions where conversions are involved.

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1. One Dimensional conversion.

Given a length of 100 inches, convert this to centimeters.

The one dimensionsl (linear) conversion factor is 1 in = 2.54 cm.

The calculation involves  100 in x 2.54 cm / in = 254 cm.

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2. Two Dimensional conversion.

Given a surface area of 500 square inches (500 in^2), convert this to square centimeters.

The two dimensional (Area) conversion factor is (1 in)^2 = (2.54 cm)^2  or 1 in^2 = 6.4516 cm^2

The calculation involves  500 in^2 x 6.4516 cm^2 / in^2 = 3,225.8 cm^2

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3. Three Dimensional conversion.

Given a volume of 1200 cubic inches (1200 in^3), convert this to cubic centimeters.

The three dimesnional (Volume) conversion factor is (1 in)^3 = (2.54 cm)^3  or  1 in^3 = 16.387064 cm^3

The calculation involves  1200 in^3 x 16.387064 cm^2 / in^3 = 19,664.4768 cm^3

Dimensional analysis focuses on the units of measurement, which is an area of mathematics that many students overlook.  Dimensional analysis focuses on the use of measurements in calculations.  Below are some examples of using units in calculations.

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1. Calculating the surface area of a shelf board.

Given a 24 inch by 48 inch shelf board calculate the amount of surface area available to store objects on.

Equation to be used:   Area = length x width

Substitution:          Area = 48 inches x 24 inches

Answer:                Area = 1,152 square inches  or  1,152 in^2

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2. Converting the surface area determined in the above problem into square feet.

The shelf board above has a surface area of  1,152 in^2

There are two ways to go about this.  One is to convert the original dimensions to feet and then substitute into the equation and solve.  The other way is to solve for the answer by doing a conversion calculation.

The 1st way:  The length is  48 in x 1 ft / 12 in = 4 ft
                       The width is  24 in x 1 ft / 12 in = 2 ft

Calculation of Area:     A = L x w = 4 ft x 2 ft = 8 square feet  or  8 ft^2

The 2nd way:  The area is  1,152 in^2

The conversion factor between the inch and the foot is  12 in = 1 ft

The conversion factor between the square inch and the square foot involves squaring both sides of the above conversion factor.  This gives us:

Conversion factor in^2 --> to ft^2 is  (12 in)^2 = (1 ft)^2

Using this conversion factor:  1,152 in^2 x (1 ft^2 / 144 in^2)

Answer:     8 ft^2

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3. The same approach is used in volume problems.

Calculate the volume of a box having the following dimensions.

     48 inches long, 24 inches wide, 36 inches deep (or high)

Equation to be used:     Volume = length x width x depth (or height)

Substitution:            Volume = 48 in  x  24 in  x  36 in

Answer:                  Volume = 41,472 in^3

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4. Determining the volume in cubic feet would involve the following process.

Again we could approach this in two different ways.

The 1st way:     The length is 48 in x 1 ft / 12 in = 4 ft
                         The width is 24 in x 1 ft / 12 in = 2 ft
                         The depth is 36 in x 1 ft / 12 in = 3 ft (also height)

Calculation of Volume:     V = L x W x H = 4 ft x 2 ft x 3 ft = 24 cubic feet  or  24 ft^3

The 2nd way:     The area is  41,472 in^3

The conversion factor between the inch and the foot is  12 in = 1 ft

The conversion factor between the cubic inch and the cubic foot involves cubing both sides of the above conversion factor.  This gives us:

Conversion factor in^3 --> to ft^3 is  (12 in)^3 = (1 ft)^3

Using this conversion factor:  41,472 in^2 x (1 ft^2 / 1,728 in^2)

Answer:     24 ft^3

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5. Determining the Density of a substance.

The mass of a rectangular solid piece of metal is  100 g

Its dimensions are:  Length is 5 cm, Width is 4 cm, and thickness (height) is 2 cm.

Equation used:     Density = Mass / Volume

Volume is:     V = L x W x H = 5 cm x 4 cm x 2 cm = 40 cm^3

Substitution:     D = M / V = 100 g / 40 cm^3 = 2.5 g/cm^3

Note:  Density is a derived unit with a complex unit of measure.

The three Trigonometry functions of Sine, Cosine, and Tangent are very important to the study of vectors in physics.  All three functions are defined as ratios of lengths of sides in a right triangle (a triangle with one angle being 90 degrees such that two of the sides of the triangle meet as lines perpendicular to each other). In such a situation the some of the other two of the three angles present in the right triangle add up to 90 degrees and are called complementary angles.

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The Sine function is described as the ratio of the length of the side opposite a defined angle (other than the 90 degree angle) to the length of the hypotenuse.

In equation form it is written as:

For example:  In a triangle having one of the angles equaling 30 degrees the ratio of the length of the side opposite to the angle to the length of the hypotenuse is 0.5000

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The Cosine function is described as the ratio of the length of the side adjacent to a defined angle (other than the 90 degree angle) to the length of the hypotenuse.

In equation form it is written as:

For example:   In a triangle having one of the angles equaling 30 degrees the ratio of the length of the side adjacent to the angle to the length of the hypotenuse is 0.8660

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The Tangent function is described as the ratio of the length of the opposite side to a defined angle (other than the 90 degree angle) to the length of the side adjacent the defined angle.

In equation form it is written as:

For example:   In a triangle having one of the angles equaling 30 degrees the the ratio of the length of the side opposite to the angle to the length of the side adjacent the angle is 0.5774