Glory to God in the highest; and on earth, peace to people on whom His favor rests! - Luke 2:14

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Fractions, Decimals, and Percents

I greet you this day,

First: Review the Notes.
Second: View the Videos.
Third: Practice questions and receive instant feedback. NO Calculators allowed.
Fourth: Solve the questions/solved examples.
Fifth: Check your solutions with my thoroughly-explained solved examples.

I wrote the codes for the calculators using JavaScript, a client-side scripting language and AJAX, a JavaScript library. Please use the latest Internet browsers. The calculators should work.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me. If you are my student, please do not contact me here. Contact me via the school's system.

Samuel Dominic Chukwuemeka (SamDom For Peace) B.Eng., A.A.T, M.Ed., M.S

Objectives

Students will:

(1.) Define natural numbers.

(2.) Define whole numbers.

(3.) Define fractions.

(4.) Define rational numbers.

(5.) Define decimals.

(6.) Define percents.

(7.) Write whole numbers as rational numbers.

(8.) Write mixed numbers as rational numbers.

(9.) Determine if a given rational number is a repeating decimal.

(10.) Determine if a given rational number is a terminating decimal.

(11.) Write fractions as decimals.

(12.) Write mixed numbers as decimals.

(13.) Write terminating decimals as fractions.

(14.) Order rational numbers from least to greatest.

(15.) Order rational numbers from greatest to least.

(16.) Solve real-world problems involving rational numbers.

(17.) Add fractions with same denominators.

(18.) Subtract fractions with same denominators.

(19.) Add mixed numbers with same denominators.

(20. Subtract mixed numbers with same denominators.

(21.) Simplify variable expressions involving "Like" fractions.

(22.) Solve linear equations in one variable involving "Like" fractions.

(23.) Check the solutions of linear equations in one variable involving "Like" fractions.

(24.) Solve real-world problems involving "Like" fractions.

(25.) Add fractions with different denominators.

(26.) Subtract fractions with different denominators.

(27.) Add mixed numbers with different denominators.

(28. Subtract mixed numbers with different denominators.

(29.) Simplify variable expressions involving "Unlike" fractions.

(30.) Solve linear equations in one variable involving "Unlike" fractions.

(31.) Check the solutions of linear equations in one variable involving "Unlike" fractions.

(32.) Solve real-world problems involving "Unlike" fractions.

(33.) Write percents as fractions.

(34.) Write fractions as percents.

(35.) Determine the percent of a given number.

(36.) Solve algebraic expressions involving percents.

(37.) Solve algebraic equations involving percents.

(38.) Solve real-world problems involving percents.

Skills Measured/Acquired

(1.) Use of prior knowledge

(2.) Critical Thinking

(3.) Interdisciplinary connections/applications

(4.) Technology

(5.) Active participation through direct questioning

(6.) Student collaboration in Final Project

Vocabulary Words

natural number, counting number, whole number, fraction, numerator, denominator, proper fraction, improper fraction, rational number, integer, mixed number, decimal, terminating decimal, exact decimal, repeating decimal, recurring decimal, irrational number, real number, complex number, add, sum, subtract, difference, multiply, product, divide, quotient, "like" fractions, "unlike" fractions, variable expressions, linear equations, equivalent fractions, same denominators, different denominators, common denominator, least common multiple (LCM)/least common denominator (LCD), proportions, equations, greatest common factor (GCF)/greatest common divisor (GCD)/highest common factor (HCF), prime factorization, left hand side (LHS), right hand side (RHS), percent, percentages, ratio, arithmetic, arithmetic operators, augend, addend, minuend, subtrahend, multiplier, multiplicand, factor, dividend, divisor, positive, negative, nonpositive, nonnegative, constant, number, variable, term, equal, equality, ratios

Definitions

A natural number is any positive integer.

It is also known as a counting number. It is a number you can count.

It does not include zero.

It does not include the negative integers.

It is not a fraction.

It is not a decimal.

A whole number is any nonnegative integer.

It includes zero and the positive integers.

It does not include the negative integers.

It is not a fraction.

It is not a decimal.

An integer is any whole number or its opposite.

Integers include the whole numbers and the negative values of the whole numbers.

(1.) The difference between a positive integer and a nonnegative integer
(2.) The difference between a negative integer and a nonpositive integer

A positive integer does not include zero but a nonnegative integer includes zero
Nonnegative integers are the positive integers and zero.

A negative integer does not include zero but a nonpositive integer includes zero
Nonpositive integers are the negative integers and zero.

A fraction is a part of a whole.

It is the part of something out of a whole thing.

It is also seen as a ratio.

It is also seen as a quotient.

The numerator is the part.

It is the "top" part of the fraction.

The denominator is the whole.

It is the "bottom" part of the fraction.

A proper fraction is a fraction whose numerator is less than the denominator.

An improper fraction is a fraction whose numerator is greater than or equal to the denominator.

A mixed number is a combination of an integer and a proper fraction.

A decimal is a linear array of digits that represent a real number, expressed in a decimal system with a decimal point; and in which every decimal place indicates a multiple of negative power of 10.

A terminating decimal is a decimal with a finite number of digits.

A terminating decimal is also known as an exact decimal.

A repeating decimal is a decimal in which one or more digits is repeated indefinitely in a pattern or sequence.

A repeating decimal is also known as a recurring decimal.

A non-repeating decimal is a decimal in which there is no sequence of repeated digits indefinitely.

A non-repeating decimal is also known as a non-recurring decimal.

A rational number is any number that can be written as a fraction where the denominator is not equal to zero.

You can also say that a rational number is a ratio of two integers where the denominator is not equal to zero.

A rational number is a number that can be written as: $${c\over d}$$ where $c, d$ are integers and $d \neq 0$

A rational number can be an integer.

It can be a terminating decimal. Why?

It can be a repeating decimal. Why?

It cannot be a non-repeating decimal. Why?

Ask students to tell you what happens if the denominator is zero.

An irrational number is a number that cannot be expressed as a fraction, terminating decimal, or repeating decimal.

When you compute irrational numbers, they are non-repeating decimals.

A real number is any rational or irrational number.

It includes all numbers that can be found on the real number line.

A complex number is a number that can be expressed in the form of $a + bi$ where $a \:and\: b$ are real numbers, and $i$ is an imaginary number equal to the square root of $-1$.

Variable Expressions are expressions that contain variables.

Linear Equations are equations in which the highest degree of the variable in the equation is 1.

Equivalent Fractions are two or more fractions that have the same value when they are expressed in their simplest forms.

Common Denominators are the common multiples of the different denominators of unlike fractions.

Least Common Denominator is the least of all the common multiples of the different denominators of unlike fractions.

A prime number is a whole number greater than $1$, which is divisible by only $1$ and itself without a remainder.

Let us look at a prime number in another way...in terms of Factors and Multiples

$Say: \\[3ex] 2 * 3 = 6 \\[3ex] 2\:\:is\:\:a\:\:factor\:\:of\:\:6 \\[3ex] 3\:\:is\:\:a\:\:factor\:\:of\:\:6 \\[3ex] 6\:\:is\:\:a\:\:multiple\:\:of\:\:2 \\[3ex] 6\:\:is\:\:a\:\:multiple\:\:of\:\:3 \\[3ex] NOTE: \\[3ex] 1\:\:is\:\:a\:\:factor\:\:of\:\:Everything \\[3ex] Everything\:\:is\:\:a\:\:multiple\:\:of\:\:1 \\[3ex]$ A prime number is a number whose factors are only $1$ and itself.
In other words, for a prime number: the only factors are $1$ and that number. No other number is a factor.

Examples include: $2, 3, 5, 7, 11, 13, 17, 19$, etc.

Prime Factorization is a method used for finding the least common denominator of unlike fractions, in which each denominator is broken down into a product of prime numbers.
This means that each denominator is split into a product of prime factors.
This means that the prime factorization of a number is the product of it's prime factors.
In other words, the prime factorization of a number is the product of it's factors that are prime numbers.

The basic arithmetic operators are the addition symbol, $+$, the subtraction symbol, $-$, the multiplication symbol, $*$, and the division symbol, $\div$

Augend is the term that is being added to. It is the first term.

Addend is the term that is added. It is the second term.

Sum is the result of the addition.

$$3 + 7 = 10$$ $$3 = augend$$ $$7 = addend$$ $$10 = sum$$

Minuend is the term that is being subtracted from. It is the first term.

Subtrahend is the term that is subtracted. It is the second term.

Difference is the result of the subtraction.

$$3 - 7 = -4$$ $$3 = minuend$$ $$7 = subtrahend$$ $$-4 = difference$$

Multiplier is the term that is multiplied by. It is the first term.

Multiplicand is the term that is multiplied. It is the second term.

Product is the result of the multiplication.

$$3 * 10 = 30$$ $$3 = multiplier$$ $$10 = multiplicand$$ $$30 = product$$

Dividend is the term that is being divided. It is the numerator.

Divisor is the term that is dividing. It is the denominator.

Quotient is the result of the division.

Remainder is the term remaining after the division.

$$12 \div 7 = 1 \:R\: 5$$ $$12 = dividend$$ $$10 = divisor$$ $$1 = quotient$$ $$5 = remainder$$

A constant is something that does not change. In mathematics, numbers are usually the constants.

A variable is something that varies (changes). In Mathematics, alphabets are usually the variables.

A mathematical expression is a combination of variables and/or constants using arithmetic operators.

A mathematical equation is an equality of two terms - the term or expression on the LHS (Left Hand Side) and the term or expression on the RHS (Right Hand Side).
This implies that we should always check the solution of any equation that we solve to make sure the LHS is equal to the RHS.
Whenever we solve for the variable in "any" equation, how do we know we are correct? CHECK!

The solution of an equation is the value of the variable which when substituted in the equation ensures that the LHS is equal to the RHS.

The solution of an equation is also known as the root of the equation or the zero of the function.

The solutions of an equation are the values of the variables which when substituted in the equation ensures that the LHS is equal to the RHS.

The solutions of an equation are also referred to as the roots of the equation or the zeros of the function.

A percent means something out of $100$.

A ratio is a comparison of two quantities.

A proportion is the equality of two ratios.

Fractions

A fraction is a part of a whole.

It is the part of something out of a whole thing.

It is also seen as a ratio.

It is also seen as a quotient.

The numerator is the part.

It is the top part of the fraction.

The denominator is the whole.

It is the bottom part of the fraction.

A proper fraction is a fraction whose numerator is less than the denominator.

An improper fraction is a fraction whose numerator is greater than or equal to the denominator.

Like Fractions are fractions with the same denominator.

Unlike Fractions are fractions with different denominators.

Examples: Introduction to Fractions

For the following scenarios, identify the common denominator.

(1.) $\dfrac{2}{3}$ of the students in Mr. C's class are females and $\dfrac{1}{3}$ are males.

Common denominator = $3$

(2.) $48$ of the $50$ states in the United States are in the continental US while $2$ of the $50$ states are not.

Common denominator = $50$

(3.) $9$ in every $10$ leaders in Nigeria are corrupt (hence the poor masses suffer terribly) while $1$ in every $10$ Nigerians are probably not.

Common denominator = $10$

(4.) Peace bought a large pizza and divided it into $12$ equal parts for her $12$ children. Each child got a part.

Common denominator = $12$

For the following scenarios, identify the different denominators.

(5.) $\dfrac{2}{5}$ of the students in Mr. C's class are females. $\dfrac{1}{3}$ of those females do not like math.

Different denominators = $5 \:and\: 3$

(6.) $9$ in every $10$ leaders in Nigeria are corrupt (hence the poor masses suffer terribly) while $1$ in every $5$ Nigerians live in less than $1 a day. Different denominators =$10 \:and\: 5$(7.) Felicity bought a large pizza and shared it among her$5$children. The first child got one-half of the pizza. The second child got one-quarter of the pizza. Different denominators =$2 \:and\: 4$(8.) Haggai bought a large pizza and shared it among her$5$children. The first child got one-half of the pizza. The second child got one-quarter of the remainder. Different denominators =$2 \:and\: 8$Add and Subtract Like Fractions To add two or more "Like Fractions": Add the numerators of the fractions. Keep the common denominator. Simplify, or convert to a mixed number as necessary. To subtract two or more Like Fractions: Subtract the numerators of the fractions. Keep the common denominator. Simplify, or convert to a mixed number as necessary. Examples: Add and Subtract "Like Fractions (1.) Add$\dfrac{3}{16} \:and\: \dfrac{11}{16} \dfrac{3}{16} + \dfrac{11}{16} \\[5ex] = \dfrac{3 + 11}{16} \\[5ex] = \dfrac{14}{16} \\[5ex] = \dfrac{7}{8} $(2.) Subtract$\dfrac{3}{16} \:from\: \dfrac{11}{16} \dfrac{11}{16} - \dfrac{3}{16} \\[5ex] = \dfrac{11 - 3}{16} \\[5ex] = \dfrac{8}{16} \\[5ex] = \dfrac{1}{2} $(3.) Subtract$\dfrac{11}{16} \:from\: \dfrac{3}{16} \dfrac{3}{16} - \dfrac{11}{16} \\[5ex] = \dfrac{3 - 11}{16} \\[5ex] = -\dfrac{8}{16} \\[5ex] = -\dfrac{1}{2} $Ask students to explain the difference between Question 2 and Question 3 based on the terminology used. (4.) Simplify$c + d - e$if$c = \dfrac{9}{61}, d = \dfrac{40}{61}, \:and\: e = \dfrac{15}{61} \dfrac{9}{61} + \dfrac{40}{61} - \dfrac{15}{61} \\[5ex] = \dfrac{9 + 40 - 15}{61} \\[5ex] = \dfrac{34}{61} $(5.) Solve and Check for the variable in the equation:$p - \dfrac{1}{14} = \dfrac{5}{14} Add\: \dfrac{1}{14} \:to\: \:both\: \:sides \\[5ex] p - \dfrac{1}{14} + \dfrac{1}{14} = \dfrac{5}{14} + \dfrac{1}{14} \\[5ex] = \dfrac{5 + 1}{14} \\[5ex] = \dfrac{6}{14} \\[5ex] = \dfrac{3}{7} $Check your work with the original equation. Left Hand Side (LHS) Right Hand Side (RHS) $$p - \dfrac{1}{14} \\[5ex] \dfrac{3}{7} - \dfrac{1}{14} \\[5ex] \dfrac{6}{14} - \dfrac{1}{14} \\[5ex] \dfrac{6 - 1}{14} \\[5ex] \dfrac{5}{14}$$ $$\dfrac{5}{14}$$ (6.) Francis completed$\dfrac{12}{28}$of his humanitarian project 2 weeks ago. Last week, he completed$\dfrac{6}{28}$of his project. What portion of his project has he not completed?$ Portion \:of\: \:his\: \:project\: \:completed\: \\[3ex] = \dfrac{12}{28} + \dfrac{6}{28} \\[5ex] = \dfrac{12 + 6}{28} \\[5ex] = \dfrac{18}{28} \\[5ex] Entire \:project\: = 1 = \dfrac{28}{28} \\[5ex] Portion \:of\: \:his\: \:project\: \:not\: \:completed\: \\[3ex] = \dfrac{28}{28} - \dfrac{18}{28} \\[5ex] = \dfrac{28 - 18}{28} \\[5ex] = \dfrac{10}{28} \\[5ex] = \dfrac{5}{14} $Do you understand this concept? Questions Answers (Click "Answer:" to show/hide answers) $$\dfrac{5}{24} + \dfrac{2}{24}$$ Answer: $$\dfrac{5}{24} - \dfrac{2}{24}$$ Answer: $$\dfrac{19a}{32} - \dfrac{27a}{32} + \dfrac{4a}{32}$$ Answer: Solve and Check for the variable in the equation: $$x - \dfrac{3}{4} = \dfrac{4}{14}$$ Answer: Solve and Check for the variable in the equation: $$\dfrac{17}{25} = k + -\dfrac{3}{25}$$ Answer: Margaret is still owing$\dfrac{7}{10}$of her auto loan. She just paid$\dfrac{3}{10}$of her loans. What fraction is still owed? Answer: Philemon purchased$\dfrac{9}{10}$acre of land in 1997 and$\dfrac{3}{10}$acre in 1998. He planted tomatoes on$\dfrac{7}{10}$acre of the land and corn on the remainder. How much land is planted in corn? Answer: St. Vincent de Paul's Charities raised$\dfrac{5}{12}$of their fundraising goal in the morning. By afternoon, they received a donation of$\dfrac{11}{12}$of their goal. By what fraction was their goal surpassed? Answer: Add and Subtract Unlike Fractions To add or subtract two or more "Unlike Fractions": Change the different denominators of the unlike fractions to a least common denominator. In other words; of all the common denominators, select the least. Convert those unlike fractions to like fractions based on their least common denominator. In other words, find the equivalent fractions based on the least common denominator. Add or subtract the numerators of the fractions as the case may be. Keep the common denominator. Simplify, or convert to a mixed number as necessary. Ask students to ask a question. What kind of question do you expect? How do we find the least common denominator of two or more unlike fractions? Least Common Denominator Least Common Denominator is the least of all the common multiples of the different denominators of unlike fractions. The least common denominator (LCD) is also referred to as the least common multiple (LCM). There are three methods of determining the LCM. (1.) Listing Multiples Method List the multiples of each of the denominators in ascending order, and stop until you get the first common multiple of both denominators. This method is usually the "more work". (2.) Prime Factorization Method Find the prime factorization of each of the denominators. Then, the least common denominator is found by multiplying all the products of the prime factors. In doing so; it is important to note that if the same number is a factor of both denominators, that number should be counted as only one factor. This method is the "most popular". (3.) Nigerian Method I learned this method in the elementary school in Nigeria. I have forgotten the name of the method. It works! It can calculate the least common multiple of two or more positive integers. In this method; we set up a table of the positive integers. We set it up the same way we set up the dividend and divisor in the Long Division method of dividing polynomials. We "try" each factor in ascending order (from least to greatest). But, if we find another factor of all the integers (a number that divides all the integers "right away" without a remainder), use it. The factor should be a positive integer that has to divide "all" the positive integers without a remainder. In other words, it has to be a common factor. If we cannot find any factor, then we divide with the smallest "given" integer. Use the "factor" to divide each positive integer. If that "factor" divides any integer, write the quotient. If that "factor" does not divide any integer, write the integer "as is" Multiply all the factors you listed. The product gives you the$LCM$. Examples: Least Common Multiple (LCM) (1.) Find the least common multiple of$3 \:and\: 15$First Method: Listing Multiples Method Multiples of$3$are 3, 6, 9, 12, 15, ... Multiples of$15$are 15, ... stop here Common multiple =$15LCM = 15$Second Method: Prime Factorization Method The prime factorization of:$3 = 315 = 3 * 53$occurred twice in$3 \:and\: 15$, so we count it only one time.$LCM = 3 * 5 = 15$Third Method: Nigerian Method $$\begin{array}{c|c c} 3 & 3 & 15 \\ \hline 5 & 1 & 5 \\ \hline & 1 & 1 \end{array}$$ The last numbers are only$1$'s STOP. The factors listed are$3$and$5$Multiply them.$LCM(3, 15) = 3 * 5 = 15$(2.) Find the least common multiple of$12 \:and\: 16$First Method: Listing Multiples Method Multiples of$12$are 12, 24, 36, 48, ... Multiples of$16$are 16, 32, 48, ... Common multiple =$48LCM = 48$Second Method: Prime Factorization Method The prime factorization of:$12 = 2 * 2 * 316 = 2 * 2 * 2 * 22$occurred twice in$12 \:and\: 16$, so we count it only one time.$2$occurred twice in$12 \:and\: 16$, so we count it only one time.$LCM = 2 * 2 * 3 * 2 * 2 = 48$Third Method: Nigerian Method $$\begin{array}{c|c c} 4 & 12 & 16 \\ \hline 3 & 3 & 4 \\ \hline 4 & 1 & 4 \\ \hline & 1 & 1 \end{array}$$ The last numbers are only$1$'s STOP. The factors listed are$4, 3$and$4$Multiply them.$LCM(12, 16) = 4 * 3 * 4 = 48$(3.) Find the least common multiple of$20 \:and\: 25$First Method: Listing Multiples Method Multiples of$20$are 20, 40, 60, 80, 100, ... Multiples of$25$are 25, 50, 75, 100, ... Common multiple =$15LCM = 15$Second Method: Prime Factorization Method The prime factorization of:$20 = 2 * 2 * 525 = 5 * 55$occurred twice in$20 \:and\: 25$, so we count it only one time.$LCM = 2 * 2 * 5 * 5 = 100$Third Method: Nigerian Method $$\begin{array}{c|c c} 5 & 20 & 25 \\ \hline 4 & 4 & 5 \\ \hline 5 & 1 & 5 \\ \hline & 1 & 1 \end{array}$$ The last numbers are only$1$'s STOP. The factors listed are$5, 4$and$5$Multiply them.$LCM(20, 25) = 5 * 4 * 5 = 100$(4.) Find the least common multiple of$5, 7, \:and\: 10$First Method: Listing Multiples Method Multiples of$5$are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, ... Multiples of$7$are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 ... Multiples of$10$are 10, 20, 30, 40, 50, 60, 70, ... Common multiple =$70LCM = 70$Second Method: Prime Factorization Method The prime factorization of:$5 = 57 = 710 = 2 * 55$occurred twice in$5 \:and\: 10$, so we count it only one time.$LCM = 5 * 7 * 2 = 70$Third Method: Nigerian Method $$\begin{array}{c|c c} 5 & 5 & 7 & 10 \\ \hline 2 & 1 & 7 & 2 \\ \hline 7 & 1 & 7 & 1 \\ \hline & 1 & 1 & 1 \end{array}$$ The last numbers are only$1$'s STOP. The factors listed are$5, 2$and$7$Multiply them.$LCM(5, 7, 10) = 5 * 2 * 7 = 70$(5.) Samuel and Dominic walk regularly to school. Samuel walks every other day. Dominic walks every fifth day. How many days will it be before they both walk on the same day? (Hint: Find the LCM of 2 and 5). First Method: Listing Multiples Method Multiples of$2$are 2, 4, 6, 8, 10, ... Multiples of$5$are 5, 10, ... Common multiple =$10LCM = 10$Second Method: Prime Factorization Method The prime factorization of:$2 = 25 = 5LCM = 2 * 5 = 10$Third Method: Nigerian Method $$\begin{array}{c|c c} 2 & 2 & 5 \\ \hline 5 & 1 & 5 \\ \hline & 1 & 1 \end{array}$$ The last numbers are only$1$'s STOP. The factors listed are$2$and$5$Multiply them.$LCM(2, 5) = 2 * 5 = 10$Ask students to say their preferred method. Do they prefer the first, second, or third method? After LCM, What Next? We have to express the denominators of the unlike fractions to have a common denominator. That common denominator is the Least Common Denominator Doing so means that we have to find an Equivalent Fraction of the unlike fraction; that has the least common denominator as its denominator. (6.) Find the equivalent fractions of$\dfrac{7}{12} \:and\: \dfrac{11}{16}$First, let us find the LCM of$12 \:and\: 16$LCM of$12 \:and\: 16 = 48\dfrac{7}{12} = \dfrac{?}{48}$In other words, what is the equivalent fraction of$\dfrac{7}{12}$that has a denominator of$48$? What do we multiply by$12$to give$48$?$12 * 4 = 48$Whatever we do to the denominator, we have to do the same thing to the numerator. If we multiply the denominator by$4$, then we also have multiply the numerator by$47 * 4 = 28$We do this to get the equivalent fraction based on the new common denominator known as the LCD (or the LCM) Therefore, the equivalent fraction of:$\dfrac{7}{12} = \dfrac{7 * 4}{12 * 4} = \dfrac{28}{48}$On to the next one:$\dfrac{11}{16}\dfrac{11}{16} = \dfrac{?}{48}$In other words, what is the equivalent fraction of$\dfrac{11}{16}$based on the LCD of$48$? What do we multiply by$16$to give$48$?$16 * 3 = 48$Whatever we do to the denominator, we have to do the same thing to the numerator. If we multiply the denominator by$3$, then we also have multiply the numerator by$311 * 3 = 33$We do this to get the equivalent fraction based on the LCD. Therefore, the equivalent fraction of:$\dfrac{11}{16} = \dfrac{11 * 3}{16 * 3} = $\dfrac{33}{48}$

We now have $\dfrac{28}{48}$ and $\dfrac{33}{48}$

These are now Like Fractions

As you can see, we have converted Unlike fractions to Like fractions
That is the intention. ☺
Whenever we have Unlike fractions, we have to convert them to Like Fractions first.
We have to find the LCD.
Then, we find the equivalent fractions based on that LCD.

(7.) Add $\dfrac{5}{12} \:\:and\:\: \dfrac{7}{16}$

$LCD \:\:of\:\: 12 \:\:and\:\: 16 = 48 \\[3ex] \dfrac{5}{12} = \dfrac{20}{48} \\[5ex] \dfrac{7}{16} = \dfrac{21}{48} \\[5ex] \dfrac{5}{12} + \dfrac{7}{16} \\[5ex] = \dfrac{20}{48} + \dfrac{21}{48} \\[5ex] = \dfrac{20 + 21}{48} \\[5ex] = \dfrac{41}{48}$

(8.) Subtract $\dfrac{3}{4} \:\:from\:\: \dfrac{9}{10}$

$LCD \:\:of\:\: 4 \:\:and\:\: 10 = 20 \\[3ex] \dfrac{9}{10} = \dfrac{18}{20} \\[5ex] \dfrac{3}{4} = \dfrac{15}{20} \\[5ex] \dfrac{9}{10} - \dfrac{3}{4} \\[5ex] = \dfrac{18}{20} - \dfrac{15}{20} \\[5ex] = \dfrac{18 - 15}{20} \\[5ex] = \dfrac{3}{20}$

Do you understand this concept?

$$\dfrac{4}{15} + \dfrac{1}{6}$$
$$\dfrac{5}{12} - \dfrac{4}{15}$$
$$\dfrac{4k}{5} - \dfrac{k}{7} + \dfrac{3k}{10}$$
Solve and Check for the variable in the equation: $$p - \dfrac{2}{9} = \dfrac{5}{12}$$
Solve and Check for the variable in the equation: $$\dfrac{5}{12} = g + -\dfrac{1}{16}$$
A teacher asked Doe how he spends his Saturdays. A day is $24$ hours.
Doe specified his time as shown in the table and on the chart.
Use the table and the chart to answer the next two questions.
Doe's Daily Time
Assignments $2$
Facebook $8$
Eat $3$
Watch TV $4$
Sleep $7$
What fraction of the day does Doe spend on eating and sleeping?
In which activity does Doe spend his greatest amount of time?
What fraction is it?
In which activity does Doe spend his least amount of time?
What fraction is it?
Is Joe a diligent student?

The figure shows a hole in a mounting bracket.

The diameter is the distance across the center of the hole.
Calculate the diameter of the hole.

A mixed number is a combination of an integer and a proper fraction.

There are two ways to add and subtract mixed numbers.

First Method

Add or Subtract the fractions first. Then, simplify as necessary.

Add the integer result with the fraction result.

Leave your answer as a mixed number, proper fraction, or integer as appropriate.

Second Method

Convert the mixed numbers to improper fractions.

Simplify, and leave your answer as a mixed number, proper fraction, or integer as appropriate.

(1.) Add $7\dfrac{1}{2}$ and $4\dfrac{1}{2}$

First Method

$Fractions:\: \dfrac{1}{2} + \dfrac{1}{2} \\[5ex] = \dfrac{1 + 1}{2} \\[5ex] = \dfrac{2}{2} \\[5ex] = 1 \\[3ex] Integers:\: 7 + 4 = 11 \\[3ex] 11 + 1 = 12$

Second Method

$7\dfrac{1}{2} = \dfrac{15}{2} \\[5ex] 4\dfrac{1}{2} = \dfrac{9}{2} \\[5ex] \dfrac{15}{2} + \dfrac{9}{2} \\[5ex] = \dfrac{15 + 9}{8} \\[5ex] = \dfrac{24}{2} \\[5ex] = 12$

(2.) Subtract $8\dfrac{3}{8}$ and $5\dfrac{1}{8}$

First Method

$Fractions:\: \dfrac{3}{8} - \dfrac{1}{8} \\[5ex] = \dfrac{3 - 1}{8} \\[5ex] = \dfrac{2}{8} \\[5ex] = \dfrac{1}{4} \\[5ex] Integers:\: 8 - 5 = 3 \\[3ex] 3 + \dfrac{1}{4} \\[5ex] = 3\dfrac{1}{4}$

Second Method

$8\dfrac{3}{8} = \dfrac{67}{8} \\[5ex] 5\dfrac{1}{8} = \dfrac{41}{8} \\[5ex] 8\dfrac{3}{8} - 5\dfrac{1}{8} \\[5ex] = \dfrac{67}{8} - \dfrac{41}{8} \\[5ex] = \dfrac{67 - 41}{8} \\[5ex] = \dfrac{26}{8} \\[5ex] = \dfrac{13}{4} \\[5ex] = 3\dfrac{1}{4}$

(3.) The perimeter of Paul's garden is $16\dfrac{1}{4}$ feet.

Paul has $14\dfrac{3}{4}$ feet of fencing material.

How many more feet does he need in order to fence the entire garden?

To solve the question, we have to subtract $14\dfrac{3}{4}$ feet from $16\dfrac{1}{4}$ feet.

That gives us the remaining amount of material needed to fence the entire garden.

$16\dfrac{1}{4} - 14\dfrac{3}{4} = ?$

First Method

$Fractions:\: \dfrac{1}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{1 - 3}{4} \\[5ex] = -\dfrac{2}{4} \\[5ex] = -\dfrac{1}{2} \\[5ex]$ This gives us a negative answer. We do not want that.
For Mixed Numbers using the First Method:
When we subtract the fractions (a bigger fraction from a smaller fraction) and get a negative number;
We have to take "1" from the integer that has the smaller fraction (the bigger integer) and
Add it to the smaller fraction to increase it's value
Then, subtract again to get a positive number.
So, take $1$ from $16$. That leaves us with $15$.
Add that $1$ to $\dfrac{1}{4}$

$1 + \dfrac{1}{4} \\[5ex] = \dfrac{4}{4} + \dfrac{1}{4} \\[5ex] = \dfrac{4 + 1}{4} \\[5ex] = \dfrac{5}{4} \\[5ex]$ Let's do it again

$Fractions:\: \dfrac{5}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{5 - 3}{4} \\[5ex] = \dfrac{2}{4} \\[5ex] = \dfrac{1}{2} \\[5ex] Integers:\: 15 - 14 = 1 \\[3ex] 1 + \dfrac{1}{2} = 1\dfrac{1}{2}$

Second Method

$16\dfrac{1}{4} = \dfrac{65}{4} \\[5ex] 14\dfrac{3}{4} = \dfrac{59}{4} \\[5ex] \dfrac{65}{4} - \dfrac{59}{4} \\[5ex] = \dfrac{65 - 59}{4} \\[5ex] = \dfrac{6}{4} \\[5ex] = \dfrac{3}{2} \\[5ex] = 1\dfrac{1}{2}$

(4.) Perform the operation: $31\dfrac{3}{5} + 18\dfrac{1}{2}$

First Method

$LCD \:of\: 5 \:and\: 2 = 10$

$Fractions:\: \dfrac{3}{5} + \dfrac{1}{2} \\[5ex] = \dfrac{6}{10} + \dfrac{5}{10} \\[5ex] = \dfrac{6 + 5}{10} \\[5ex] = \dfrac{11}{10} \\[5ex] = 1\dfrac{1}{10} \\[5ex] Integers:\: 31 + 18 = 49 \\[3ex] Updated\:Integers:\: 31 + 18 + 1 = 50 \\[3ex] 50 + \dfrac{1}{10} = 50\dfrac{1}{10}$

Second Method

$LCD \:of\: 5 \:and\: 2 = 10 \\[3ex] 31\dfrac{3}{5} = \dfrac{158}{5} \\[5ex] 18\dfrac{1}{2} = \dfrac{37}{2} \\[5ex] \dfrac{158}{5} + \dfrac{37}{2} \\[5ex] = \dfrac{316}{10} + \dfrac{185}{10} \\[5ex] = \dfrac{316 +185}{10} \\[5ex] = \dfrac{501}{10} \\[5ex] = 50\dfrac{1}{10}$

(5.) Perform the operation: $6\dfrac{2}{5} - 4\dfrac{7}{10}$

First Method

$LCD \:of\: 5 \:and\: 10 = 10 \\[3ex] Fractions:\: \dfrac{2}{5} - \dfrac{7}{10} \\[5ex] = \dfrac{4}{10} - \dfrac{7}{10} \\[5ex] = \dfrac{4 - 7}{10} \\[5ex] = -\dfrac{3}{10} \\[5ex]$ We do not want that
So, we need to borrow $1$ from $6$, and add to $\dfrac{2}{5}$

Borrow $1$ from $6$: $6 - 1 = 5$

Add to $\dfrac{2}{5}$

$1 + \dfrac{2}{5} \\[5ex] = \dfrac{5}{5} + \dfrac{2}{5} \\[5ex] = \dfrac{5 + 2}{5} \\[5ex] = \dfrac{7}{5} \\[5ex]$ We do the subtraction again.

$Fractions:\: \dfrac{7}{5} - \dfrac{7}{10} \\[5ex] = \dfrac{14}{10} - \dfrac{7}{10} \\[5ex] = \dfrac{14 - 7}{10} \\[5ex] = \dfrac{7}{10} \\[5ex] Integers: 5 - 4 = 1 \\[3ex] 1 + \dfrac{7}{10} \\[5ex] = \dfrac{10}{10} + \dfrac{7}{10} \\[5ex] = \dfrac{10 + 7}{10} \\[5ex] = \dfrac{17}{10} \\[5ex] = 1\dfrac{7}{10} \\[5ex]$ Some students may ask you this question:
Why not just do it this way: $1 + \dfrac{7}{10} = 1\dfrac{7}{10}$?

Why do it the longer way? Be prepared to explain. Let them know they can do it either way.
When you add an integer and a fraction, just write/combine it as a mixed number.
Please do not do it when you subtract, multiply, or divide.
You only do it when you add.

Second Method

$LCD \:of\: 5 \:\:and\:\: 10 = 10 \\[3ex] 6\dfrac{2}{5} = \dfrac{32}{5} \\[5ex] 4\dfrac{7}{10} = \dfrac{47}{10} \\[5ex] \dfrac{32}{5} - \dfrac{47}{10} \\[5ex] = \dfrac{64}{10} - \dfrac{47}{10} \\[5ex] = \dfrac{64 - 47}{10} \\[5ex] = \dfrac{17}{10} \\[5ex] = 1\dfrac{7}{10}$

(6.) Evaluate the expression: $c - d$ if $c = 6$ and $d = -3\dfrac{1}{2}$

First Method

$c - d \\[3ex] 6 - \left(-3\dfrac{1}{2}\right) \\[5ex] 6 + 3\dfrac{1}{2} \\[5ex] Fractions:\: \dfrac{1}{2} \\[5ex] Integers:\: 6 + 3 = 9 \\[3ex] 9 + \dfrac{1}{2} = 9\dfrac{1}{2}$

Second Method
$c - d \\[3ex] 6 - \left(-3\dfrac{1}{2}\right) \\[5ex] 6 + 3\dfrac{1}{2} \\[5ex] 6 = \dfrac{6}{1} \\[5ex] LCD \:of\: 1 \:and\: 2 = 2 \\[3ex] 3\dfrac{1}{2} = \dfrac{7}{2} \\[5ex] \dfrac{6}{1} + \dfrac{7}{2} \\[5ex] = \dfrac{12}{2} + \dfrac{7}{2} \\[5ex] = \dfrac{12 + 7}{2} \\[5ex] = \dfrac{19}{2} \\[5ex] = 9\dfrac{1}{2}$

Ask students to state their preferences between the first and second methods. They should give reasons for their preferences.

(7.) Solve and Check the equation for the variable.

$3\dfrac{1}{4} + 2\dfrac{1}{6} = 5\dfrac{5}{12} + y$

First Method

Rearrange: Make the LHS (Left Hand Side) to be the RHS (Right Hand Side), and vice versa.

$y + 5\dfrac{5}{12} = 3\dfrac{1}{4} + 2\dfrac{1}{6} \\[5ex] y = 3\dfrac{1}{4} + 2\dfrac{1}{6} - 5\dfrac{5}{12} \\[5ex] y = \dfrac{13}{4} + \dfrac{13}{6} - \dfrac{65}{12} \\[5ex] LCD \:of\: 4, 6, \:and\: 12 = 12 \\[3ex] y = \dfrac{39}{12} + \dfrac{26}{12} - \dfrac{65}{12} \\[5ex] = \dfrac{39 + 26 - 65}{12} \\[5ex] = \dfrac{65 - 65}{12} \\[5ex] = \dfrac{0}{12} \\[5ex] = 0 \\[3ex] y = 0$

Second Method
Can we do this another way?
Whenever you have any equation involving fractions, you can convert all the "terms" to integers by multiplying each term by the LCD
You do this if you do not want to work with fractions. ☺
But, you have to convert all mixed numbers to improper fractions before you use this method.
Rearrange: Make the LHS (Left Hand Side) to be the RHS (Right Hand Side), and vice versa.

$y + 5\dfrac{5}{12} = 3\dfrac{1}{4} + 2\dfrac{1}{6} \\[5ex] y + \dfrac{65}{12} = \dfrac{13}{4} + \dfrac{13}{6} \\[5ex] LCD \:of\: 4, 6, \:and\: 12 = 12 \\[3ex] Multiply\:\:each\:\:term\:\:by\:\:the\:\:LCD \\[3ex] 12 * y + 12 * \dfrac{65}{12} = 12 * \dfrac{13}{4} + 12 * \dfrac{13}{6} \\[5ex] 12y + 65 = 3 * 13 + 2 * 13 \\[3ex] We\:\:now\:\:have\:\:only\:\:integers \\[3ex] 12y + 65 = 39 + 26 \\[3ex] Subtract\:\:65\:\:from\:\:both\:\:sides \\[3ex] 12y = 39 + 26 - 65 \\[3ex] 12y = 0 \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:12 \\[3ex] y = 0$

Check your work. Check with the original equation.

Left Hand Side (LHS) Right Hand Side (RHS)

$$3\dfrac{1}{4} + 2\dfrac{1}{6} \\[5ex] \dfrac{13}{4} + \dfrac{13}{6} \\[5ex] \dfrac{39}{12} - \dfrac{26}{12} \\[5ex] \dfrac{39 + 26}{12} \\[5ex] \dfrac{65}{12} \\[5ex] 5\dfrac{5}{12}$$

$$5\dfrac{5}{12} + y \\[5ex] 5\dfrac{5}{12} + 0 \\[5ex] 5\dfrac{5}{12}$$

Ask students why we need to "always check equations".
Some might actually ask you first.
If they do, seek their thoughts first.

(8.) James and John went to fish.

James caught a fish that was $18\dfrac{1}{4}$ inches long.

John caught a fish that was $15\dfrac{1}{2}$ inches long.

How much longer is James's fish than John's?

We have to subtract $15\dfrac{1}{2} inches \:from\: 18\dfrac{1}{4} inches$

That gives us the length by which James' fish is longer than John's.

$18\dfrac{1}{4} - 15\dfrac{1}{2} \\[5ex] 18\dfrac{1}{4} = \dfrac{73}{4} \\[5ex] 15\dfrac{1}{2} = \dfrac{31}{2} \\[5ex] LCD \:of\: 4 \:and\: 2 = 4 \\[3ex] \dfrac{73}{4} - \dfrac{31}{2} \\[5ex] = \dfrac{73}{4} - \dfrac{62}{4} \\[5ex] = \dfrac{73 - 62}{4} \\[5ex] = \dfrac{11}{4} \\[5ex] = 2\dfrac{3}{4}\:inches \\[5ex]$ This means that James' fish is $2\dfrac{3}{4} inches$ longer than John's.

Do you understand this concept?

$$3\dfrac{1}{2} + 9\dfrac{1}{2}$$
$$-3\dfrac{1}{8} + 8\dfrac{3}{8}$$
$$2\dfrac{3}{4} + 3\dfrac{1}{5}$$
$$-2\dfrac{3}{4} + 3\dfrac{1}{5}$$

Percents

How have you used the word, "per"?
Say: miles per hour (mph) means $\dfrac{miles}{hour}$

Per indicates a division

A percent means something out of $100$.
"per" indicates a division.
"cent" is a hundred.
"Per cent" means "Per hundred"

$3\%$ means $3$ out of $100$ = $\dfrac{3}{100}$

$70\%$ means $70$ out of $100$ = $\dfrac{70}{100} = \dfrac{7}{10}$

$Samdom\%$ means $Samdom$ out of $100$ = $\dfrac{Samdom}{100}$

$c\%$ means $c$ out of $100$ = $\dfrac{c}{100}$

So, whenever we talk of percent; we are comparing something to $100$

We can also see "Percent" as a:
(1.) Ratio: This is because we compare something to $100$
Because it is a comparison, it is also a ratio.
That "something" can be a constant or variable.

(2.) Fraction: This is because when we say $21\%$;

it means $\dfrac{21}{100}$

This is a fraction.
$21$ is the part
$100$ is the whole

Scenarios Related to Percents
(1.) My mother was very proud of me because I earned a $\boldsymbol{100\%}$ on my mathematics test!
(2.) The National Weather Service predicted a $\boldsymbol{25\%}$ chance of snow this weekend.
(3.) President Barack Obama won $\boldsymbol{71\%}$ of the Latino votes in the $2012$ Presidential elections in the United States.

What other ways have you used the term, percent?

Percent Application Formulas

Percent of Change
(1.) Change = New - Initial
(2.) Increase: Change is positive. $New \gt Initial$
(3.) Decrease: Change is negative. $New \lt Initial$

(4.) $\%\:\: of\:\: Change = \dfrac{Change}{Initial} * 100$

(5.) $\%\:\: of\:\: Increase = \dfrac{New - Initial}{Initial} * 100$

(5.) $\%\:\: of\:\: Decrease = \dfrac{Initial - New}{Initial} * 100$

NOTE: We do not include the negative sign for the Percent of Decrease.
We already know it is a decrease.

Percentage Error
(1.) Error = Measured - Actual

(2.) $\% Error = \dfrac{Measured - Actual}{Actual} * 100$

Percents and Proportions
(1.) $\dfrac{is}{of} = \dfrac{\%}{100}$

This is also the same as writing that:

$\dfrac{part}{whole} = \dfrac{\%}{100}$

Percent Equations
(1.) "is" means "$=$"
(2.) "of" means "multiply"
(3.) $\%$ means "out of $100$"
(4.) "what" means the variable.

Profit, Loss, $\%$ Profit, $\%$ Loss, Sales, Discounts

$(1.)\:\: Profit = Selling\:\:Price - Cost\:\:Price \\[3ex] (2.)\:\: Loss = Cost\:\:Price - Selling\:\:Price \\[3ex] (3.)\:\: \%Profit = \dfrac{Profit}{Cost\:\:Price} * 100 \\[5ex] (4.)\:\: \%Loss = \dfrac{Loss}{Cost\:\:Price} * 100 \\[5ex] (5.)\:\: Sale\:\:Price = Initial\:\:Price - Discount \\[3ex] (6.)\:\: Discount = Initial\:\:Price - Sale\:\:Price \\[3ex] (7.)\:\: \%Discount = \dfrac{Discount}{Initial\:\:Price} * 100 \\[5ex]$

Practice with That Quiz

Practice what you just learned. Calculators are NOT allowed.

(1.) Visit the website:

(2.) Under Fractions, begin with Identify, then do Simplify, Inequality, and Arithmetic

(3.) Set the Length to $50$

(4.) Set the Level to $50$

(5.) Attempt all checkboxes and/or radiobuttons

(6.) Go back to (4.) and set the Level to $100$

Rounding Decimals Calculator

References

Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com

Bittinger, M. L., Beecher, J. A., Ellenbogen, D. J., & Penna, J. A. (2017). Algebra and Trigonometry: Graphs and Models ($6^{th}$ ed.). Boston: Pearson.

Coburn, J., & Coffelt, J. (2014). College Algebra Essentials ($3^{rd}$ ed.). New York: McGraw-Hill

David Sobecki, P., & Mercer, B. A. (2016). Math in Our World: A Quantitative Reasoning Approach. New York, NY: McGraw-Hill Education.

Kaufmann, J., & Schwitters, K. (2011). Algebra for College Students (Revised/Expanded ed.). Belmont, CA: Brooks/Cole, Cengage Learning.

Lial, M., & Hornsby, J. (2012). Beginning and Intermediate Algebra (Revised/Expanded ed.). Boston: Pearson Addison-Wesley.