Package Summary - Yola · Solving Linear Equations Percentages Let’s Play! (Exercises) [email protected] 1 of 17 c Sa’diyya Hendrickson. Understanding Variables Level:

Introduction Prepared by Sa’diyya Hendrickson

Name: Date:

Package Summary

• Understanding Variables

• Translations

• The Distributive Property

• Expanding Expressions

• Collecting Like Terms

• Solving Linear Equations

• Percentages

• Let’s Play! (Exercises)

[email protected] 1 of 17 c© Sa’diyya Hendrickson

Understanding Variables Level: X

• variable: a variable is a letter that is used to: (1) represent a quantity that we

don’t know or (2) be a placeholder for the input of one or more numbers.

Because variables are just numbers in disguise, everything that we’ve learned about num-

bers still applies! Let’s look at some examples:


Understanding Variables Level: X

1. area: Area is a property of two-dimensional figures and is the number of 1 × 1squares needed to cover the figure. Recall that the area of a rectangle is given by:length × width since this product produces the number of 1× 1 squares needed tocover the rectangle.

2. volume: Volume is a property of three-dimensional figures and is the number of1 × 1 × 1 cubes needed to fill the same amount of space as the figure. Recall thatthe volume of a rectangular prism is given by: length × width × height .


Translations Level: X

Below are partial statements, with bolded terms that re-

quire us to perform one of following five operations : ad-

dition, subtraction, multiplication, division, or ex-

ponentiation. List each term under the correct

category.

1. five more toys than Jasmine

2. snacks shared evenly between four people

3. seven less than Tristan’s mark

4. four times the width

5. the square of the length

6. pencils distributed to eight students

7. the product of John’s age and Kayley’s age

8. one-third of Yan’s age

9. rent increased by $50.00

10. triple the cost of the shirt

11. four centimetres taller than Kamar

12. five years younger than Hamsha

13. $10 off of the regular price

14. twice as many markers

15. the distance cubed

16. the quotient of two numbers

17. the difference between Sam’s and Samir’s height


Translations Level: X

After identifying appropriate operations (if any), it’s useful to translate English statements

to simpler, algebraic expressions. First, name the variable using a “let” statement!

Examples:

1. Find an expression for two consecutive integers.

Solution: The word consecutive means “following continuously” or “in a row.” So,

to name two consecutive integers, we simply let n = any integer. Then, n + 1 must

be the integer that follows (because integers differ by 1)!

2. Find an expression for a tuition that has been increased by $500.

Solution: If we let t = previous tuition costs, then t+ $500 represents the new total.

Q: Why do we take the time to name a variable before writing the expression, rather

than just creating an expression with the actual word(s)?

e.g. (any integer)+1 or (previous tuition)+$500

A: It’s more efficient to write and perform operations on simpler expressions!


Distributive Property Level: X

The distributive property has come up time and time again, which really highlightshow important it is. The property states: a(b + c) = ab + ac for integers a, b, and c

a) factored to expanded form b) expanded to factored form

Let’s use area models to better understand this property. If a, b and c are positive integerssuch that a and b + c represent the length and width of a rectangle, then the area ofthe rectangle is given by:

A = l · w = a (b + c)

We can also look at this in a different way. The area of this rectangle is just the sum ofthe areas of two smaller rectangles!


Distributive Property Level: X

1. Expand the expression: 3(2x + 4)

3(2x + 4) = 3(2x) + 3(4) by the distributive property

= (3 · 2)x + 12 by the associative property of multiplication

= 6x + 12

2. Factor the expression: 8x + 14

8x + 14 = (2 · 4)x + (2 · 7) numbers are even ⇒ common factor of 2

= 2 (4x) + 2 (7) by the associative property of multiplication

= 2(4x + 7) factoring out 2 by the distributive property

• Type 1: distributivity over subtraction (i.e. distributing over negative values)

a(b− c) = ab− ac

We will prove why the distributive property generalizes over subtraction (i.e. positive

and negative integers) after we’ve learned how to solve equations!

• Case 2: distributing negative values inside of the brackets

−a(b + c) = −ab− ac

Consider the justification below:

−a(b + c) = (−1)(a)(b + c) by definition of multiplying by −1

= (−1)(a(b + c)) by the associative property of multiplication

= −(ab + ac) by the distributive property

= the opposite of ab + ac by definition of opposites

= −ab + (−ac) since the opposite of ab + ac is:

(the opposite of ab) plus (the opposite of ac)

= −ab− ac by definition of subtraction as adding the opposite


Expanding Expressions Level: X

Expand the following expressions:

1. 7(4x2 − 5)

7(4x2 − 5) = 7(4x2)− 7(5) by the distributive property

= (7 · 4)x2 − 35 by the associative property of multiplication

= 28x2 − 35

2. −2x(−3x3 + 9x)

−2x(−3x3 + 9x) = (−2x)(−3x3) + (−2x)(9x) by the distributive property

= (−2)(−3)(x3 · x1) + (−2)(9)(x · x) by the associative property

= 6 + ( )x2 by properties of exponents

= by the definition of subtraction

3. −4x(−6x2 + x− 8)

−4x(−6x2 + x− 8) = (−4x)(−6x2) + (−4x)(x)− (−4x)(8) by the distributive property

= (−4)(−6)(x1 · x2) + (−4)(x · x)− (−4)(8)x by the associative property

= x3 + (−4)x2 − (−32)x by properties of exponents

= by definition of subtraction

Can you identify the solutions using the distributive property and mental math?

1. 6x(2x− 3) (a) 24x2 − 8x− 40

2. − 7x2(−2x2 + x− 1) (b) 12x2 + 18

3. 8(−3x2 − x + 5) (c) 14x4 + 7x3 + 7x2

4. − 6x(−2x− 3) (d) − 24x2 − 8x + 40

5. 7x2(2x2 + x + 1) (e) 12x2 − 18x

6. 8(3x2 − x− 5) (f) 14x4 − 7x3 + 7x2


Collecting Like Terms Level: X

Suppose we are asked to expand an expression that yields the following result:

4x2 − 4 + 2x + x2 − 5x3 + 7− x3 − x

The terms in an expression are separated by addition symbols (all subtraction symbols

can be rewritten so that we are adding the opposite). The original expression becomes:

4x2 + (−4) + 2x + x2 + (−5x3) + 7 + (−x3) + (−x)

Above, there are terms: 4x2, −4, 2x, , −5x3, 7, , and −x

It’s all About the Powers

Simplifying an expression requires that we group terms based on their powers. Given

some variable, say x, we have like terms if they belong to the same “powers family.”

1. The Constants (x0):

e.g. 1, −2, 5, etc.

2. First family (x1):

e.g. x, −2x, 5x, etc.

3. Squared family (x2):

e.g. x2, −2x2, 5x2, etc.

4. Cubed family (x3):

e.g. x3, −2x3, 5x3, etc.

and so on . . .

How Many do We Have?

To simplify, we add/subtract terms that belong to the same family! For example, in the

Cubed family we have: five negative x3s and one positive x3. So, one of the negatives will

cancel with the positive, leaving behind four negative x3s (i.e. −4x3). Algebraically:

−5x3 + x3 = −5 x3 + x3 x3 is a common factor

= (−5 + 1)x3 by the distributive property

= −4x3

So, simplifying: (−5x3 + x3) + (4x2 + x2) + (2x− x) + (−4 + 7) grouping families

= (−5 + 1)x3 + (4 + 1)x2 + (2− 1)x + 3 by the distributive property

= −4x3 + 5x2 + x + 3


Equations Level: X

An equation is a mathematical statement involving an equal sign: a = b

The equal sign separates the left side (L.S.) of the equation from the right

side (R.S.). Below are some important properties of equations.

Let a, b and c be any numbers. Then:

1. Reflexive Property: If a = b, then b = a e.g. If x + 3 = 5, then 5 = x + 3.

2. If a = b, then a± c = b± c

In other words, as long as we add or subtract the same thing on both sides, the

equality still holds!

3. If a = b, then a · c = b · cIn other words, as long as we multiply by the same thing on both sides, the equality

still holds!

4. If a = b and c 6= 0, then ac

= bc

In other words, as long as we divide by the same thing on both sides, the equality

still holds!

Q: Suppose we were asked to find a number such that when you double it and then

increase the result by 1, the answer is 5. How would you find this number?

A: One way is to work backwards from 5! We must undo everything that was done to

create the number 5 (i.e. do the opposite)!

• First, subtract the 1 that was added to increase the result to 5. This returns 4.

• Then, take the number 4 and divide it by 2, to undo the doubling (i.e. multiplication

by 2). So, the answer is 2.


Equations Level: X

The previous question was simple enough that we could work backwards fairly easily to

find the desired number. However, if the problem becomes more complicated, this is not

an efficient method. Let’s consider using an algebraic expression:

Let x = {some number} such that: 2x + 1 = 5.

Below, we will do the opposite to both sides to balance the equation and transform

2x + 1 = 5 into x = {some number}


Equations Level: X

Summarizing our Strategies:

• S1 Identify which side of the equation x is on. If there is more than one x,

decide which side you want to “move” all of them to.

• S2 Do the opposite to rearrange the equation so that x is on one side (by itself)

and numbers are on the other side.

– Addition and are opposite operations!

– Multiplication and are opposite operations!

• S3 Sneak up on x! The ultimate goal is to isolate x, and the best approach is to

get everything out of x’s way, starting from the numbers furthest away!

Finally, we check that our solution is correct. In other words, we need to make sure that

when x = −12, L.S. = R.S.

L.S. = 4− x

= 4−(−1

2

)=

4 · 21 · 2

+1

2

=9

2

R.S. = 5x + 7

= 5

(−1

2

)+ 7

= −5

2+

7 · 21 · 2

=9

2

Therefore, L.S. = R.S.


Equations Level: X

Solve the same equation: 4 − x = 5x + 7 by moving the x’s to the R.S. You should get

the same answer!

Proving the Distributive Property over Subtraction

Now we have the skills to prove the distributive property over subtraction:

a(b− c) = ab− ac

Proof: First, we let k = b− c.

k = b− c ⇒ k +c = b− c +c adding c to both sides

⇒ k + c = b since −c + c = 0

⇒ a (k + c) = a (b) multiplying both sides by a

⇒ ak + ac = ab by the distributive property on L.S.

⇒ ak + ac −ac = ab −ac subtracting ac from both sides

⇒ ak = ab− ac since ac− ac = 0

⇒ a(b− c) = ab− ac since k = b− c


Percents Level: X

• percent:

percent comes from the latin phrase per centum and means “per hundred.”

In other words: p % =p

100

– Example 1: 33 % means “33 per hundred,” which is just the fraction 33100

.

– Example 2: When we purchase an item from a store and are charged 13 % in

taxes, we are paying 113 % of the retail price. This is because we are paying

100 % of the retail price plus an additional 13 % in taxes.

Rounding

When asked to round a decimal number to a nearest place value, the convention is to:

• Look at the place value immediately to the right, and . . .

– If the digit in that position is greater than or equal to 5 (i.e. 5− 9), round up

– If the digit in that position is less than 5 (i.e. 0− 4), we round down

Example: Round 45.47683 % to the nearest hundredth.

• First we identify the place value to the right of the hundredths place

(i.e. ). Let’s underline it: 45.4 7 683.

• Because the number is 6 (which is greater than 5), we round 7 up to 8, resulting in:

45.48 %.

Let’s explore this in detail on the next page!


Percents Level: X

We have the following four cases:

I. Percents to Fractions

II. Percents to Decimals

III. Decimals to Percents

IV. Fractions to Percents

I. Percents to Fractions

By the definition of percent, we simply drop the % symbol write the value over 100!

e.g. 26% =26

100=

13

50in reduced form

II. Percents to Decimals

Recall that: p% = p100

. In the previous lesson, we learned that whenever we divide a

number by 10k, we simply move the decimal point k places to the left ! Here, we have:

100 = 102.

Strategies: e.g. Express 435 % as a decimal

• Drop the % symbol: e.g. 435

• Identify where the decimal is (if there isn’t one, place one at the end): e.g. 435.

• Then move the decimal two places to the left ! e.g. 4.35

III. Decimals to Percents

To go from percents to decimals, we have to divide by 100, so if we want to go from

decimals to percents, we simply have to do the opposite (i.e. multiply by 100)!

In the previous lesson, we learned that whenever we multiply a number by 10k, we simply

move the decimal point k places to the right ! Here, we have: 100 = 102.

Strategies: e.g. Express 0.345 as a percent

• Identify where the decimal is (if there isn’t one, place one at the end): e.g. 0.345

(the decimal is between the 0 and the 3)

• Move the decimal two places to the right ! e.g. 34.5

• Place the % symbol at the end: e.g. 34.5 %


Percents Level: X

IV. Fractions to Percents

To go from fractions to percents, we must satisfy the following equation:

a

b= p%, which is equivalent to

a

b=

p

100(where a and b are integers)

If the fraction already has a denominator of 100, there is nothing to do! Otherwise, we

simply need to see that above, we have an equation where we have to solve for p! Notice

that to isolate p, we simply need to do the opposite of dividing by 100

(i.e. by 100)!

Strategies: e.g. Express 58

as a percent

• Multiply the fraction by 100 and find the equivalent decimal:

p =5

8· 100

1=

5

2· 25

1by reducing the pair (8, 100)

=125

2by definition of fraction multiplication

= 62.5 by long division or mental math

• Place the % symbol at the end: e.g. 62.5 %

1. A store is offering 30% off of a $75 item that you’d like to purchase. How much will

you save?

• S1: Write the percent as a decimal i.e. 30% = 0.3

• S2: Translate the English statement into an algebraic equation

i.e. Let s = amount saved. Then s = 0.3 · 75 since “of” means to multiply here!

• S3: Calculate i.e. s = 0.3 · 75 = 22.5, meaning that you save $22.50

2. Calculate the 20% tip of a $15 dine-in meal using mental math.

• First notice that you get 20% by doubling 10%.

• Now we can use the fact that: 10% = 10100

= 110

, so “10% of 15” translates to

“ 110

(15)” = 1510

= 1.5 (since dividing by 10 means moving the decimal point one

place to the left ! Doubling $1.50 produces a tip of $3!


Let’s Play! Level: X

1. Give translations for the English statements on page 4. Be sure to begin

with the appropriate “let” statements.

2. Use the distributive property to expand the following expressions:

(a) 5(x + 4) (b) 2(6x + 5) (c) x(8x + 2)

(d) 6x(2x + 3) (e) x(3x2 + 4) (f) 2(x2 + 3x + 4)

(g) 7x(2x2 + x + 1) (h) x(3x2 + 4x + 9) (i) 2x(4x2 + 5x + 7)

3. Simplify the following expressions (i.e. first expand if necessary (using

the distributive property) and then collect like terms):

(a) 2x− 5x (b) x− (3x + 5)

(c) 4x + 3− (x + 6) (d) 2x2 − x− x2(3x + 4)

(e) x3 + 2x− 5− x(x2 − x + 4) (f) −5x4 − x3 + 2x2 + x2(x2 − 3x− 2)

4. Solve the following equations:

(a) 4x + 2 = 14 (b) 3− 2x = −11 (c) 3x− 8 = −17

(d) 5− x = 3x− 9 (e) x− 4 = 5− 7x (f) 2x + 9 = 7 + 10x

5. Find the fraction, decimal and/or percent representations:

(a) 120

(b) 3.57 (c) 76.3%

(d) 254

(e) 42.8 (f) 45%

6. Suppose you wanted to purchase a new winter jacket that has a retail price of $300.

(a) Calculate how much additional money you need to cover the taxes (13%).

(b) If you received $170 on your birthday to put towards this purchase, what percent-

age of the purchase price has been covered? (be sure to create a “let” statement

for the unknown percentage, and then translate the english statement: “what

percent of $300 equals $170” into an equation that you can solve.)


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Package Summary - Yola · Solving Linear Equations Percentages Let’s Play! (Exercises) [email protected] 1 of 17 c Sa’diyya Hendrickson. Understanding Variables Level: