UNIT 2 - EQUATIONS EQUATIONS A mathematical statement in which two quantities are equal is an

equation. Mathematical shorthand is used in equations using numbers, letters and operational symbols. Unknowns or variables are letters that represent unknown amounts.

Known or given amounts are represented by numbers. The mathematical symbols, numbers and letters show us how the knowns are related, for example in solving an equation like 10 = 2 B, means to find the value of B so that 2 times an unknown value is the same as 10. By performing systematic operations so that the unknown value is isolated, we will make the letter representing the unknown stand alone on one side of the equation.

Solving equations using multiplication or division Looking at equations that involve multiplication or division and one unknown

value is where we will begin our examinations of equations. Solving an equation with multiplication or division

Isolating the unknown: (a.) Divide both sides of the equation by the known factor if the equation

contains the product of the unknown factor and a known factor (b.) Multiply both sides of the equation by the divisor if the equation contains

the quotient fo the unknown and the divisor. Solve the equation: 4N = 16 Testing: 4(4) = 16 4N / 4 = 16 / 4 16 = 16 N = 4

The solution is the number on the side opposite the isolated unknown-value letter Check the solution by replacing the unknown letter in the equation with the solution; perform the

indicated operations and verify that both sides of the equation are the same number.

Example Solve the equation: 5A = 30 The product and one factor are unknown

5A = 30 Divide the known factor on both sides of the equation. 5A / 5 = 30 / 5 The solution is 6

A = 6 Checking solution: Replace A with the solution (6) to see if both sides are equal 5A = 30

5(6) = 30

30 = 30 The solution of the equation is 6.

Example 2 Solve the equation: B / 7 = 4 The quotient and divisor are known. The dividend is unknown B / 7= 4 Multiply both sides of the equation by the divisor, 7. 7( B / 7)

= 7(4) The solution is 28

B = 28 Check the solution Replace A with the solution (28) and see if both sides are equal B / 7 = 4

28 / 7 = 4

4 = 4 The solution of the equation is 28

Solving equations using addition or subtraction If 23 of the 33 people work at AIL company on the day shift, how many of them

work in the evening? Since we know that 23 people work there during the day and that the total is 33 people in all, the number of people that work in the evening is the unknown. Assign a letter to the unknown number of night-shift workers in an equation.

X + 23 = 33 This equation is one that can be solved with subtraction. Solving an equation with addition or subtraction

Isolating the unknown:

(a.) Subtract both sides of the equation by the known factor if the equation contains the sum of the unknown factor and a known factor

(b.) Add on both sides of the equation by the known value if the equation contains the difference of the unknown and a known factor

Example 1 Solve the equation: X + 23 = 33 -23 -23 X = 10 Testing the answer = 10+23 = 33

=33 = 33

The solution is the number on the side opposite the isolated unknown-value letter

Check the solution by replacing the unknown letter in the equation with the solution; perform the indicated operations and verify that both sides of the equation are the same number.

Example 2 Solve the equation: A – 10 = 17 The difference and the number being subtracted are known. A - 10 = 17 Add 10 to both sides of the equation .

+10 +10 The solution is 27

A = 27 Check the solution Replace A with the solution (27) and see if both sides are equal A – 10 = 17

27 – 10 = 17

17 = 17 The solution of the equation is 27

Solving equations using more than one operation Sometimes equations will contain more than one operation and each operation will have to be undone in

turn. First undo all additions and subtractions and then undo all multiplications and divisions, but the goal is still to isolate the unknown.

Isolating the unknown:

(a.) Add or subtract as necessary (b.) Multiply or divide as necessary Solve the equation: 3N – 6 = 9 (+6 and -6 cancel each other out) +6 +6 3N = 15 3N / 3 = 15 / 3 N = 5

The solution is the number on the side opposite the isolated unknown-value letter Check the solution by replacing the unknown-value letter in the equation with the solution;

perform the indicated operations

Example 1 Solve the equation: 2A - 1 = 13. Since the equation contains both addition and multiplication,

undo addition first and then undo multiplication Undo Addition 2A - 1 = 13 +1 +1 2A = 14 Undo Multiplication 2A = 14 2A / 2 = 14 / 2 A = 7 Check your solution: Replace A with 7 in the original equation and see if both sides are equal 2A - 1 = 13 Multiply first

2(7) - 1 = 13 Add

14 - 1 = 13

13 = 13

Example 2 Solve the equation: (B / 4) – 3 = 2. Since the equation contains both subtraction and division, undo subtraction first and then undo

division Undo subtraction (B / 4) – 3 = 2

+3 +3 Undo division

B / 4 = 5

4(B / 4) = 4(5)

B= 20 Checking solution: Replace A with 20 in the original equation and see if both sides are equal (B / 4) – 3 = 2 Divide first

(20 / 4) – 3 = 2 Subtract 5 –3 = 2

2 = 2 The solution of the equation is 20

Solving equations containing multiple unknown terms The unknown value may occur more than once in some equations. When the unknown value

occurs in two addends, this is the simplest instance and we solve these types of equations by combining these addends first. For instance, 5A means 5 times A, so to combine 2A and 3A, we simply add 2 and 3, then multiply 5 by A with the result of 5A. Therefore, 2A + 3A = 5A.

Solving an equation when the unknown value occurs in two addends Solve the equation: 3C + 2C = 20 1. Combine the unknown-value addends that are on the same side of the equation (a.) Add the numbers in each addend

(3+ 2)C = 20 (b.) Represent the multiplication of their sum by the unknown value 5C = 20 2. Solve the resulting equation

5C / 5 = 20 / 5

C = 4

Example 1

Solve the equation: 4x + 3x - 1 = 20 First combine the unknown-value addends.

4x + 3x - 1 = 20 Undo subtraction

7x– 1 = 20

+1 +1 Undo multiplication

7x = 21

7x / 7 = 21 / 7x = 3

CALCULATE THE FUTURE VALUE AND THE COMPOUND INTEREST AMOUNT USING A FORMULA

Solving equations that contain parentheses First you will need to write the equation in the form that contains no parentheses Solve the equation: 2(3M - 1) = 10

Eliminate the parentheses (a.) Multiply the number outside of the parentheses by each addend inside (b.) Show the resulting products as addition ot subtraction as indicated

2(3M - 1) = 10

6M- 2 = 10 2. Solve the resulting equation 6M - 2 = 10 + 2 +2

6M = 12 6M / 6 = 12 / 6 or M = 2

FINDING THE EFFECTIVE INTEREST RATE ( APY)

Example 1 Solve the equation: 4(N + 3) = 20 First eliminate the parentheses. Multiply the number outside of the parentheses by each addend

inside, then show the products as addition 4(N + 3) = 20

4N + 12 = 20 Undo addition

- 12 -12

4N = 8 Undo multiplication

4N / 4 = 8/ 4

N = 2

Solving equations that are proportions Based on two pairs of related quantities, a proportion is most commonly written as a

fraction notation. A ratio is a number written in fraction notation is called a ratio. A proportion is formed when two fractions or ratios are equal.

It is important that the cross products are equal and the product of the numerator times the denominator forms the cross product. In the following proportion ½ = 2/4, one cross product is 1 x 4 and the other cross product is 2 x 2, notice that the two cross products are both equal to 4. Take a look at other proportions.

5/10 = 2/4 2(10) = 4(5) 20 = 20 7/14 = 1/2 2(7) = 1(14) 14 = 14 Verifying that two fractions form a proportion

find the two cross products 1/3 = 2/6 Compare the two cross products 1(6) = 6 and 2(3) = 6 If the cross products are equal, the two fractions form a proportion since 6 =6 then 1/3 and 2/6

form a proportion.

Example 1 Which fraction is proportional to 12/15, 3/4 or 4/5? Find the cross products 4/5 = 12/15 Multiply 4(15) = 5(12) 60 = 60

They are equal so 4/5 and 12/15 are a proportion Find the cross products 3/4 = 12/15 Multiply 3(15) = 4(12) 45 = 48

They are not equal so it is not a proportion

Solving a proportion

Find the cross products Undo the multiplication and isolate the unknown

Example 2 Solve the proportion: 3/5 = 9 / B Find the cross products 3/5 = 9 / B Multiply 3B = 5(9) Undo multiplication 3B = 45 Divide 3B / 3 = 45 / 3 N = 15

Using equations to solve problems Because equations use mathematical shorthand for expressing relationships, they are

powerful business tools and developing a solution plan is a critical step. Using the problem-solving approach to analyze and solve word problems There are always key words in a problem that indicate whether to add to, subtract from,

multiply or divde some number, so if a word problem states that Sarah' salary in 2008 exceeds her 2007 salary, you know that you should add $2,500 to her 2007 salary to find her 2008 salary. Most times you will see the word of in a problem and this indicates that the problem will involve multiplication. In the following table 2-1, you will find the key words and what they will generally mean when they are used in a word problem.

Example 1 At Matt’s restaurant, full-time employees work more hours per day than part-time employees. With the

difference of working 7 hours per day and part-timers work 5 hours a day, how many hours do full-timers work per day?

Solution Plan: The word difference implies subtraction. N is number of full-timers.

Full-time hours – part-time hours = difference of hours

N – 5 = 7 Solution: Undo subtraction N – 5 = 7 +5 +5 N = 12 Check the solution: Replace N with 12 and subtract 12 – 5 = 7 The sides are equal 7 = 7

Example 2 If Wanda wants to save 1/5 of her $450 wages per week. How much will she be saving each week? Solution plan: The word of implies multiplication X Amount to be saved x = 1/5 ($450) Solution: Reduce and multiply x = 1/5 ($450) The solution is 35 x = $90 Check the solution: Replace x with $90 and see if the sides are equal $90 = 1/5($450) $90 = $90 There will be times that a problem will require finding more than one unknown value and the strategy is to choose

a letter to represent one unknown value. By using the known facts, we can then express all of the other unknown values in terms of the one letter. Meaning if we know that twice as many boys than girls will attend a concert, then we might represent the number of girls as G and the number of boys as 2G, which is twice as many as G.

CALCULATE THE PRESENT VALUE USING A FORMULA

Example 3 Last Monday at Jack’s restaurant, there were twice as many requests for seats on the patio than requests for seats at the

bar. If 141 customers total came into the restaurant that day, how many requested the patio and how many requested the bar?

Solution plan: Patio + Bar = total customers ; x = number of people in the Patio P + 2P = 141 Combine like terms x + 2x = 141 3x = 141 Divide both sides by 3 3x / 3 = 141 / 3 The solution is 114 x = 47 Patio patrons 2x = 2(47) 2x = 94 number of Bar patrons is 94 Check the solution: Substitute x = 47 and 2x = 94 x + 2x = 141 47 + 94 = 141 141 = 141 Many problems will give a total number of two types of items, but you want to know the number of each of the items. See

the next example that illustrates this type of problem.

Example 4 Bob's shop spent a total of $95 ordering 60 cards from Pops Card Co., whose birthday cards cost $1.75 each

and nature cards that cost $1.50 each. How many of each kind of card did Bob's order? Solution plan: Total cost = (cost per birthday card)(number of birthday cards) + (cost per nature card)

(number of nature cards) N = number of birthday cards 95 = (1.75)(N) + (1.50)(60 – N) Solution: Eliminate parentheses showing grouping95 = 1.75N + 1.50(60 – N) Multiply 1.50(60) 95 = 1.75N+ (1.50)(60) – 1.50N Combine letter terms 95 =

1.75N + 90 – 1.50N

-90 = -90

5 = 0.25N Divide both sides by 0.25 5 = 0.25N

The solution is 20, representing the Birthday cards N = 20 Subtract 20 from 60 to find 60 – B, or 40, the number of nature cards

60 – B = 60 – 20 = 40

Bob's Shop ordered 20 birthday cards and 40 nature cards Daily problems often involve two pairs of values that are proportional. Example 5 How far can you go on 8 gallons of gas if you car gets 32 miles to a gallon? Miles traveled per 8 gallons is proportional to miles traveled for each 1 gallon. 1 gallon / 32 miles = 8 gallons / A miles

Solution: A = total number of miles Cross multiply 1/32 = 8/A Multiply 1A = (8)(32) A = 256

Example 6 To mix concentrated solution A, the container states to mix 3 ounces of solution A with every 2

gallons of water, so for 7 gallons of water, how many ounces of solution A should you use? Solution plan: Amount of solution 1 per 5 gallons is proportional to an amount of solution 1 for

each 2 gallons. 2 gallons / 3 ounces = 7 gallons / x ounces Solution: Cross multiply 2/3 = 7/x Multiply 2x = (3)(7) Divide both sides by 2 2x = 21 2x = 21 2 = 2 The solution is 7 ½ x = 10 ½

Formulas Evaluating formulas Procedures that have been used so frequently to solve certain types of problems are formulas

and have become the accepted means of solving these problems. Composed of numbers, letters, or variables, formulas are used to represent unknown numbers and operations that relate to these unknown and known values. To substitute known values for the appropriate letters is to evaluate a formula, you will then perform the indicated operations to find the unknown value. Sometimes you must first solve the equation to isolate the unknown value in the formula.

Evaluating a formula

Write the formula Substitute known values for the letters of the furmula Applying the order of operations, solve the equation for the unknown letter Interpret the solution within the context of the formula.

Example 1 Jen’s Shop purchases a TV for $615 then marks it up $200. How much will the TV

sell for? Use the following formula to solve. (S = selling cost, C = cost, M = markup) S = C + M

Write the formula and substitute known values for C and M S = C + M Add

S = $615 + $200

S = $815 The TV will sell for $815 The missing value is not always the value that is isolated in the formula in some

instances. Use the techniques for solving equations to find the missing value after the known values are substituted into the formula.

Example 2 What is the mark up on a DVD player that costs $70 and sells for $110?

Use the following formula to solve. (S = selling cost, C = cost, M = markup) S = C + M

Write the formula and substitute known values for C and S S = C

+ M Subtract $85 from each side

$110 = $70 + M

-70 -$70

$40 = M The DVD player was marked up $40.

Finding a variation of a formula by rearranging the formula Sometimes a formula can have as many variations as there are letters or variables in

the formula. Any missing number can be found no matter what it appears in formula using the techniques for solving equations. When the variation is used frequently, variations of formulas are desirable, but also in an electronic spreadsheet, the missing number should be isolated on the left side of the equation. To solve for that variable, is to isolate a variable.

Finding a variation of a formula by rearranging the formula

Isolate the variable that is to be found Treat all other variables of the formula as you would treat a number in an

equation and perform normal steps for solving equations Interchange the sides of the variable if it is isolated ont eh right side of the

equation, so that it appears on the left.

Example 1 Solve the following formula: S = C + M for C Isolate C. Subtract M from both sides of the equation S = C + M Simplify. M – M = 0. C + 0 = C S –

M = C + M – M Interchange the sides of the equation S –

M = C Formula variation C =

S – M When comparing prices of a product available in different quantities the unit price

of the product is used. The formula used is U = P / N, where U is the unit price of a specified amount of the product, P is the total price and N is the number of specified units contained in the product. Identified in many ways, the specified unit can be a measuring unit such as pounds (lb) or ounces (oz) or the number of items such as an individual snack cake in a package of cakes.

Example 2 Solve for X using a variation of the formula M = X / N Isolate X. Multiply both sides fo the equation by N

M = X / N Simplify (multiply both sides by N)

N(M) = N (X / N) Interchange the sides of the equation NM = X Formula solved for X X = NM

Discussion?