Basic Algebra Concepts: Lesson 1

November 20, 2024

Table of Contents

Basic Algebra Concepts: Lesson 1

Core Mathematical Operations

Understanding the basic operations in algebra is essential. Here’s a breakdown:

Multiplication: Repeated addition of a number.
- Example: $3 \times 4 = 12$
Addition: Summing up numbers.
- Example: $5 + 3 = 8$
Division: Splitting into equal parts.
- Example: $10 \div 2 = 5$
Subtraction: Finding the difference.
- Example: $7 – 4 = 3$

Key Algebraic Rules

Zero Rules:
- Division by zero ( $\frac{x}{0}$ ) is undefined.
- Zero divided by any number ( $\frac{0}{x}$ ) equals 0.
Order of Operations (BODMAS): The order of operations ensures expressions are solved systematically:
- B: Brackets first.
- O: Orders (Exponents and Roots).
- D: Division (from left to right).
- M: Multiplication (from left to right).
- A: Addition (from left to right).
- S: Subtraction (from left to right).
Example: Solve $3 + 6 \div 2 \times (4 – 1)$ :
$Step 1: (4 – 1) = 3 \quad \text{(Brackets first)}.$ $Step 2: 6 \div 2 = 3 \quad \text{(Division next)}.$ $Step 3: 3 \times 3 = 9 \quad \text{(Multiplication)}.$ $Step 4: 3 + 9 = 12 \quad \text{(Addition last)}.$

Example: Solving a Polynomial

Multiply $(n + 3)(n + 5)$ :

$(n + 3)(n + 5) = n^2 + 5n + 3n + 15$

Simplify:

$n^2 + 8n + 15$

Laws of Exponents

Multiplication of Exponents (Same Base):
$x^m \cdot x^n = x^{m+n}$
Division of Exponents (Same Base):
$\frac{x^m}{x^n} = x^{m-n}, \quad x \neq 0$
Power of a Power:
$(x^m)^n = x^{m \cdot n}$
Zero Exponent:
$x^0 = 1, \quad x \neq 0$
Adding Like Terms:
$ax^m + bx^m = x^m (a + b)$
Addition of Different Powers:
$x^m + x^n = x^m + x^n, \quad \text{cannot simplify further unless $m = n$}.$

Square Roots (Radicals)

Square Root Definition:
- $\sqrt{x} = x^{1/2}$
Rules for Square Roots:
- $\sqrt{x} \cdot \sqrt{y} = \sqrt{x \cdot y}$
- $\sqrt{x + y} \neq \sqrt{x} + \sqrt{y}$
- $a \sqrt{x} + b \sqrt{x} = \sqrt{x}(a + b)$
- $\frac{1}{\sqrt{x}} = x^{-1/2}$

Place Value

Place Value	Digit
Ten thousands	8
Thousands	3
Hundreds	9
Tens	0
Units/Ones	6

Example: For the number 83,906, the place values are assigned as shown.

Translating Words and Phrases into Algebraic Symbols

Word or Phrase	Symbol	Written Example	Algebraic Result
“Is” or “is equal to”	$=$	The sum of an unknown number and 3 is 5.	$x + 3 = 5$
“Sum”	$+$	The sum of two numbers is 5.	$x + y = 5$
“Difference”	$–$	How much more is 7 than -3?	$7 – (-3) = 10$
“Product”	$\cdot$	What is the product of 5 and 4?	$5 \cdot 4 = 20$
“Divided”	$\div$	What is 6 divided by 2?	$6 \div 2 = 3$
“Of” (Multiplication)	$\cdot$	5% of 100.	$0.05 \cdot 100 = 5$
“Decreased by”	$–$	A budget of $10.00 decreased by 5%.	$10 – (0.05 \cdot 10) = 9.50$
“Added” or “Increase”	$+$	By what must I increase 5 to equal 7?	$5 + x = 7$ , $x = 2$
“How many fit”	$\div$	How many 4 cubic inch boxes can fit into a 16 cubic inch box?	$16 \div 4 = 4$

Examples with Completed Calculations

Sum Example:
- The sum of two numbers is 5.
- Algebraic Expression: $x + y = 5$
- If $x = 2$ , $y = 3$ .
Difference Example:
- How much more is 7 than -3?
- Calculation: $7 – (-3) = 10$ .
Product Example:
- What is the product of 5 and 4?
- Calculation: $5 \cdot 4 = 20$ .
Division Example:
- What is 6 divided by 2?
- Calculation: $6 \div 2 = 3$ .
Percentage Example:
- A budget of $10.00 is decreased by 5%.
- Calculation: $10 – (0.05 \cdot 10) = 9.50$ .
Fit Example:
- How many 4 cubic inch boxes can fit into a 16 cubic inch box?
- Calculation: $16 \div 4 = 4$ .
Absolute Value Example:
- What must be added to 9 to equal 14?
- Calculation: $9 + x = 14$ , $x = 5$ .

Examples with Alternative Formatting

Example 1: Determining Tank Capacity

Problem: A storage tank is

\frac{3}{4}

full. After removing 5 gallons, it will be

\frac{5}{9}

full. How many gallons does the tank hold when completely full?

Step 1: Define the Variables

Let $c$ represent the total capacity of the tank in gallons.

Step 2: Establish the Equation

The amount of liquid in the tank initially is $\frac{3}{4} c$ .
After 5 gallons are removed, the remaining liquid is $\frac{5}{9} c$ .
Equation: $\frac{3}{4} c – 5 = \frac{5}{9} c$

Step 3: Solve for

c

Rearrange the equation to group like terms: $\frac{3}{4} c – \frac{5}{9} c = 5$
Find a common denominator for the fractions ( $36$ : $\frac{27}{36} c – \frac{20}{36} c = 5$
Simplify the left-hand side: $\frac{7}{36} c = 5$
Multiply through by $36$ to isolate $c$ : $c = 5 \times \frac{36}{7} = \frac{180}{7} \approx 25.71$

Step 4: Verify the Answer

Initially: $\frac{3}{4} c = \frac{3}{4} \times 25.71 \approx 19.28 \, \text{gallons}$
After removing 5 gallons: $19.28 – 5 \approx 14.28 \, \text{gallons}$
Check the fraction: $\frac{14.28}{25.71} \approx \frac{5}{9}$

Answer: The total capacity of the tank is approximately 25.71 gallons.

Example 2: Consecutive Odd Numbers

Problem: The sum of three consecutive odd numbers is 99. Find the largest number.

Step 1: Define the Variable

Let the first odd number be $n$ .
The second odd number is $n + 2$ .
The third odd number is $n + 4$ .

Step 2: Create the Equation

n + (n + 2) + (n + 4) = 99

Step 3: Solve for

n

Simplify the equation: $3n + 6 = 99$
Subtract 6 from both sides: $3n = 93$
Divide by 3: $n = 31$

Step 4: Identify the Numbers

First number: $n = 31$
Second number: $n + 2 = 33$
Third number: $n + 4 = 35$

Step 5: Verify the Solution

31 + 33 + 35 = 99

Answer: The largest of the three numbers is 35.

Example III: Dividing Money Among Michael, Paul, and Susan

Problem: $72 is to be divided among Michael, Paul, and Susan such that:

Michael receives 5 times as much as Susan.
Paul receives 3 times as much as Susan.

How much does Paul receive?

Step 1: Define Variables

Let Susan’s share be $S$ .
Michael’s share is $5S$ .
Paul’s share is $3S$ .

Step 2: Set Up the Equation The total money distributed is $72:

$S + 5S + 3S = 72$

Step 3: Solve the Equation

Combine like terms:
$9S = 72$
Solve for $S$ :
$S = \frac{72}{9} = 8$

Step 4: Find Paul’s Share Paul receives $3S$ :

$3 \times 8 = 24$

Answer:
Paul receives $24.

Note:

Michael receives $5S = 5 \times 8 = 40$ .
Susan receives $S = 8$ .
While Michael’s share is not part of the question, we now know he gets $40.

Example IV: Age Puzzle

Problem: Six years ago, I was half the age I will be in six years. How old am I now?

Step 1: Define Variables

Let my current age be $x$ .
My age six years ago: $x – 6$ .
My age six years from now: $x + 6$ .

Step 2: Set Up the Equation Six years ago, my age was half of what it will be six years from now:

$x – 6 = \frac{1}{2}(x + 6)$

Step 3: Solve the Equation

Eliminate the fraction by multiplying through by 2:
$2(x – 6) = x + 6$
Simplify:
$2x – 12 = x + 6$
Rearrange terms:
$2x – x = 6 + 12$
Solve for $x$ :
$x = 18$

Step 4: Verify the Solution

Six years ago, my age was $x – 6 = 18 – 6 = 12$ .
Six years from now, my age will be $x + 6 = 18 + 6 = 24$ .
Half of 24 is indeed 12.

Answer:
I am currently 18 years old.

1. A woman is paid $c$ dollars per hour for every hour she works up to 8 hours and is paid double for every hour she works after 8 hours. How many dollars will she be paid for working 13 hours?

(A) $13c$
(B) $\frac{13}{c}$
(C) $c$
(D) $18c$
(E) $\frac{18}{c}$

2. A number is multiplied by another number. The product is then divided by the difference between the two numbers. What is the result?

(A) $\frac{xy}{x-y}$
(B) $\frac{x+y}{x-y}$
(C) $x-y$
(D) $xy$
(E) $\frac{x-y}{x+y}$

3. Bill had $x$ dollars, and he bought $y$ apples for 16 cents and $z$ pears for 12 cents each. How many cents did he have left?

(A) $x – 16y + 12z$
(B) $100x – 6y + 12z$
(C) $x – 16y – 12z$
(D) $x – (y + z)$
(E) $100x – 16y – 12z$

4. If 3 is subtracted from a certain number, the result is 6 more than twice the number. Find the number.

(A) $-3$
(B) $-9$
(C) $3$
(D) $\frac{9}{7}$
(E) $-\frac{9}{7}$

5. How many 29-cent stamps can be purchased for $d$ dollars?

(A) $29d$
(B) $\frac{d}{29}$
(C) $\frac{29}{d}$
(D) $\frac{100d}{29}$
(E) $\frac{29}{100d}$

This matches the original problem layout and aligns the multiple-choice options under each question.

For problems 6-20, enter your solutions into the grid provided below each question.

6. On a given day in February, the temperature in a town ranged from -12 to 19 degrees. What is the difference between the high and the low temperature for the day?
Answer: The difference is the absolute value of $19 – (-12) = 19 + 12 = 31$

7. Four years ago, my age was half of what it will be in eight years. How old am I?
Let $x$ be my current age.

x – 4 = \frac{1}{2}(x + 8)

Solve:

2(x – 4) = x + 8 \implies 2x – 8 = x + 8 \implies x = 16

Answer: 16 years old.

8. Chris and Andrea have $100.00 together. If Chris gives Andrea $10.00, she will have 20 more dollars than he will. How many dollars did Chris start with?
Let $C$ be the amount Chris starts with.

C + (100 – C – 10) + 10 = C – 15

Solve: $C$ .

9. One quarter of the students at a high school take algebra. One fifth take geometry. The remaining 110 students do not take any math. How many students are there at the school?
Solve (Full Solving Mode).

Mat.

…

Here’s the corrected and formatted version of problems 6–20 with answers:

Problems 6–20: Enter your solutions into the grid provided below each question.

6. On a given day in February, the temperature in a town ranged from -12 to 19 degrees. What is the difference between the high and low temperature for the day?
Solution:
The difference is:

19 – (-12) = 19 + 12 = 31 \, \text{degrees.}

Answer: 31

7. Four years ago, my age was half of what it will be in eight years. How old am I?
Solution:
Let $x$ be my current age.

x – 4 = \frac{1}{2}(x + 8)

2(x – 4) = x + 8 \implies 2x – 8 = x + 8 \implies x = 16

Answer: 16 years old

8. Chris and Andrea have $100.00 together. If Chris gives Andrea $10.00, she will have $20 more than he will. How much money did Chris start with?
Solution:
Let $C$ be the amount Chris has.
Then Andrea has $100 – C$ .
After Chris gives $10:

(100 – C + 10) = C – 10 + 20

Solve:

100 – C + 10 = C + 10 \implies 120 = 2C \implies C = 60

Answer: Chris started with $60.

9. One quarter of the students at a high school take algebra. One fifth of the students take geometry. The remaining 110 students do not take any math. How many students are at the school?
Solution:
Let the total number of students be $x$ .

\frac{1}{4}x + \frac{1}{5}x + 110 = x

Combine fractions:

\frac{5x + 4x}{20} + 110 = x \implies \frac{9x}{20} + 110 = x

Solve:

110 = x – \frac{9x}{20} \implies 110 = \frac{20x – 9x}{20} \implies 110 = \frac{11x}{20}

x = 110 \times \frac{20}{11} = 200

Answer: 200 students

10. The sum of five consecutive integers is equal to three times the largest. What is the largest of the integers?
Solution:
Let the integers be $x-2, x-1, x, x+1, x+2$
Their sum is:

(x-2) + (x-1) + x + (x+1) + (x+2) = 5x

The largest integer is $x+2$ , and the equation is:

5x = 3(x+2) \implies 5x = 3x + 6 \implies 2x = 6 \implies x = 3

Largest integer: $x+2 = 5$ .
Answer: 5

11. A man goes to a bank with $4.00 and asks for change. He is given an equal number of nickels, dimes, and quarters. How many of each is he given?
Solution:
Let the number of each coin be $n$ .
Value equation:

5n + 10n + 25n = 400 \implies 40n = 400 \implies n = 10

Answer: He is given 10 of each coin.

12. The difference between the squares of two numbers is 9. The difference between the two numbers is 1. What is their sum?
Solution:
Let the numbers be $x$ and $y$ .
Equations:

x^2 – y^2 = 9 \quad \text{and} \quad x – y = 1

Factorize the first:

(x+y)(x-y) = 9

Substitute $x-y = 1$ :

(x+y)(1) = 9 \implies x+y = 9

Answer: Their sum is 9.

13. The volume of a box is 24 cubic inches. If its length is 3 inches and its width is 8 inches, what is its depth?
Solution:
Volume equation:

\text{Volume} = \text{length} \times \text{width} \times \text{depth}

24 = 3 \times 8 \times \text{depth} \implies 24 = 24 \cdot \text{depth} \implies \text{depth} = 1

Answer: 1 inch

14. One-third the sum of 13 and a certain number is the same as one more than twice the number. Find the number.
Solution:
Let the number be $x$ .

\frac{1}{3}(13 + x) = 1 + 2x

Clear fractions:

13 + x = 3 + 6x \implies 10 = 5x \implies x = 2

Answer: 2

15. Tom has a brother one-third his age and a sister three times his age. If the combined ages of all three is five less than twice the oldest, how old is Tom?
Solution:
Let Tom’s age be $x$ .
Brother’s age: $\frac{x}{3}$
Sister’s age: $3x$

x + \frac{x}{3} + 3x = 2(3x) – 5

Solve:

x + \frac{x}{3} + 3x = 6x – 5 \implies \frac{13x}{3} = 6x – 5

13x = 18x – 15 \implies 15 = 5x \implies x = 3

Answer: 3 years old

16. A woman leaves one-fourth of her estate to her son and one-third to her daughter. If she leaves $1000 to charity, how large was her estate?
Solution:
Let the estate be $x$ .

\frac{1}{4}x + \frac{1}{3}x + 1000 = x

Combine fractions:

\frac{3x}{12} + \frac{4x}{12} + 1000 = x \implies \frac{7x}{12} + 1000 = x

Solve:

1000 = \frac{5x}{12} \implies x = \frac{12 \cdot 1000}{5} = 2400

Answer: $2400

17. A woman buys a pound of steak for $3. If it loses one-fourth of its weight when cooked, what is the cost per pound when served?
Solution:
Remaining weight: $1 – \frac{1}{4} = \frac{3}{4}$ .
Cost per pound:

\frac{3}{\frac{3}{4}} = 4

Answer: $4 per pound

18. A gas tank is $\frac{1}{3}$ full and requires 6 gallons to make it $\frac{5}{6}$ full. What is the capacity of the tank?
Solution:
Let the capacity be $x$ .

\frac{5}{6}x – \frac{1}{3}x = 6

Simplify:

\frac{5x – 2x}{6} = 6 \implies \frac{3x}{6} = 6 \implies x = 12

Answer: 12 gallons

19. The sum of five consecutive odd integers exceeds three times the largest by 6. Find the sum of the integers.
Solution:
Let the integers be $x-4, x-2, x, x+2, x+4$ .

(x-4) + (x-2) + x + (x+2) + (x+4) = 5x

Largest: $x+4$ .

5x = 3(x+4) + 6 \implies 5x = 3x + 12 + 6 \implies 2x = 18 \implies x = 9

Sum: $5x = 45$ .
Answer: 45

20. Ten houses line one side of a street. The average space between houses is 60 feet more than the average width of each house. A sidewalk starts 60 feet before the first house and ends 60 feet after the last house. If the total length of the sidewalk is 3206 feet, find the average width of each house.
Solution:
Let the average width of each house be $w$ .
Total space:

10w + 9(w+60) + 120 = 3206

Simplify:

19w + 540 = 3206 \implies 19w = 2666 \implies w = 140

Answer: 140 feet

Understanding Inequalities and Their Manipulations

An equality is represented by an = sign, while an inequality is represented by signs such as < (less than) or > (greater than).

Examples:

$3 < 5$ : Three is less than five.
$6 > 0$ : Six is greater than zero.

If you’re unsure which direction the inequality points:

The smaller end points to the smaller number.
The open end points to the larger number.

Just like equations, inequalities can be manipulated to find solutions. However, there are some additional rules to consider, especially when the sign direction changes.

Rules for Solving Inequalities

1. Adding or Subtracting a Number

Adding or subtracting a number (positive or negative) from both sides of an inequality does not change the direction of the inequality sign.

Example:

-4 < 2 \quad \Rightarrow \quad -4 + 5 < 2 + 5 \quad \Rightarrow \quad 1 < 7

2. Multiplication or Division by a Positive Number

Multiplying or dividing both sides of an inequality by a positive number does not change the direction of the inequality sign.

Example:

1 < 3 \quad \Rightarrow \quad 4 \cdot 1 < 4 \cdot 3 \quad \Rightarrow \quad 4 < 12

3. Multiplication or Division by a Negative Number

When multiplying or dividing both sides of an inequality by a negative number, you must reverse the direction of the inequality sign.

Example:

3 < 4 \quad \Rightarrow \quad 3 \cdot (-2) > 4 \cdot (-2) \quad \Rightarrow \quad -6 > -8

4. Taking the Square Roots

Taking the square root of both sides does not change the direction of the inequality sign, but you can only take the square root of positive numbers.

Example:

4 < 5 \quad \Rightarrow \quad \sqrt{4} < \sqrt{5} \quad \Rightarrow \quad 2 < \sqrt{5}

5. Squaring Both Sides

If both sides are positive: Squaring does not change the direction.
Example:
$4 > 3 \quad \Rightarrow \quad 4^2 > 3^2 \quad \Rightarrow \quad 16 > 9$
If both sides are negative: Squaring reverses the inequality sign.
Example:
$-3 > -4 \quad \Rightarrow \quad (-3)^2 < (-4)^2 \quad \Rightarrow \quad 9 < 16$
If one side is positive and the other is negative: The inequality direction may or may not change.

6. Inverting Both Sides

Inverting both sides of an inequality reverses the direction if both sides are positive or negative.

Example:

\frac{1}{3} < \frac{3}{4} \quad \Rightarrow \quad \frac{3}{4} > \frac{4}{3}

7. Absolute Values

If $|x| > a$ (where $a > 0$ ):
$x > a \quad \text{or} \quad x < -a$
Example:
$|x| > 5 \quad \Rightarrow \quad x > 5 \quad \text{or} \quad x < -5$
If $|x| < a$ (where $a > 0$ ):
$-a < x < a$
Example:
$|x| < 7 \quad \Rightarrow \quad -7 < x < 7$

Additional Inequality Properties

If $a < b$ and $x < y$ , then $a + x < b + y$ .
Example:
$1 > -1 \quad \text{and} \quad 3 > 2 \quad \Rightarrow \quad 1 + 3 > -1 + 2 \quad \Rightarrow \quad 4 > 1$
If $x > r$ and $r > y$ , then $x > y$ .
Example:
$x > 3 \quad \text{and} \quad 3 > y \quad \Rightarrow \quad x > y$

Example: Comparing Values with Inequalities

Problem:

If $abc < 0$ and $c < 0$ , which is larger: $c$ or the product $a \cdot b$ ?

Solution:

Start with the given conditions:
$c < 0$
(Given in the problem)
$abc < 0$
(Given in the problem)
Divide both sides of $abc < 0$ by $c$ :
$\frac{abc}{c} > \frac{0}{c}$
Change the direction of the inequality:
Since $c$ is negative, dividing by it reverses the inequality direction:
$ab > 0$
Interpret the result:
$ab$ is greater than 0, and $c$ is less than 0. Therefore, $ab$ is greater than $c$ .

Special Inequality Symbols

$\leq$ : Less than or equal to.
$\geq$ : Greater than or equal to.

When solving inequalities involving these symbols, follow the same rules as for $<$ and $>$ .

Examples with $\leq$ and $\geq$

Example 1: Combining Inequalities

If $x \leq 3$ and $x \geq 3$ , then $x = 3$ because both statements are true when $x$ equals 3.

Example 2: Contradictory Inequalities

If $x < 3$ and $x > 3$ , there is no possible value for $x$ that satisfies both conditions.
Result: No solution.

Example 3: True and False Comparisons

$2 \leq 2$ : True statement.
$2 < 2$ : False statement.

Example Problems: Inequalities and Logical Reasoning

Problem 1: Solve for $x$

Equation:

5x – 3 < 12

Solution:

Add 3 to both sides:
$5x – 3 + 3 < 12 + 3$ $5x < 15$
Divide both sides by 5:
$x < 3$

Final Answer:

x < 3

Problem 2: Logical Reasoning

Question:
If $x + y = z$ and $y > 0$ , which of the following statements cannot be true?

(A) $x > z$
(B) $x + y > -z$
(C) $y = z$
(D) $x + z > 1$
(E) $z > x + 1$

Solution (Forward Method):

Given $y > 0$ , multiply both sides by -1 and flip the inequality:
$-y < 0$
Add $z$ to both sides:
$z – y < z + 0$
Substitute $x + y$ for $z$ :
$x + y – y < z$ $x < z$

Since $x < z$ , statement (A): $x > z$ is false.

Solution (Backward Method):

Start with the statement $x > z$ :
Add $y$ to both sides:
$x + y > z + y$
Subtract $z$ from both sides:
$x + y – z > z + y – z$ $(x + y) – (x + y) > y$ $0 > y$

This contradicts the given condition $y > 0$ , so statement (A): $x > z$ is false.

Important Note:

One common mistake is to infer too much from an inequality.

Example: If $a < b$ , this does not imply:

That $a$ is negative and $b$ is positive.
Both $a$ and $b$ could be negative or positive.

The inequality only specifies their relative values, not their signs.

Practice Problems with Solutions and Explanations:

Problem 1:

If $x – 3y > x + 3y$ , then which of the following must be false?

Solution:
Simplify the inequality:

$x – 3y > x + 3y$

Subtract $x$ from both sides:

$-3y > 3y$

Divide both sides by 3:

$-1 > y$

This means $y < -1$ .

Answer: (A) $y > 0$ (must be false).

Problem 2:

If $a^2 – 2ab + b^2 > a^2 + 2ab + b^2$ and $b < 0$ , which of the following describes $a$ ?

Solution:
Simplify the inequality:

$a^2 – 2ab + b^2 > a^2 + 2ab + b^2$

Cancel $a^2$ and $b^2$ :

$-2ab > 2ab$

Combine terms:

$-4ab > 0$

Since $b < 0$ , divide both sides by $-4b$ (flip the inequality):

$a < 0$

Answer: (C) $a < 0$ .

Problem 3:

If $p \geq 0$ , then:

(A) $-p < -3p$
(B) $-2p > 0$
(C) $7p \leq 4p$
(D) $6p \geq 2p$
(E) None of the above

Solution:
Evaluate each option:

$-p < -3p$ : False when $p \geq 0$ .
$-2p > 0$ : False when $p \geq 0$ .
$7p \leq 4p$ : False when $p > 0$ .
$6p \geq 2p$ : True when $p \geq 0$ .

Answer: (D) $6p \geq 2p$ .

Problem 4:

If $a > b$ , and $b > c$ , then:

(A) $a > 0$
(B) $c > 0$
(C) $a – b > b – c$
(D) $abc < 0$
(E) None of the above

Solution:
Let’s test with $a = 3$ , $b = 2$ , $c = 1$ :

$a > b$ and $b > c$ are true.
$a – b > b – c$ : False, $3 – 2 \neq 2 – 1$ .
$abc < 0$ : False since $abc > 0$ .

Answer: (E) None of the above.

Problem 5:

If $a > b$ and $b > c$ , then:

(A) $4a > 3b$
(B) $2a > 0$
(C) $a – b > 0$
(D) $ab < 0$
(E) $a^2 > b^2$

Solution:
From $a > b$ and $b > c$ :

$a – b > 0$ : True.

Answer: (C) $a – b > 0$ .

Problem 6:

If $a < -1$ , which statement is not true?

(A) $a + 4 < 3$
(B) $-3a < 3$
(C) $a – 5 < -6$
(D) $5a < -5$
(E) $-4a > 4$

Solution:
From $a < -1$ :

$-3a < 3$ : True since dividing $-3a$ flips inequality.

Answer: (B) $-3a < 3$ (not true).

Problem 7:

If $x < y$ and $y < -4$ , which statement is not true?

(A) $-4 < x$
(B) $x – 2 < y – 2$
(C) $x < -4$
(D) $2x < -8$
(E) $x + y < -4$

Solution:
From $x < y$ and $y < -4$ :

$-4 < x$ : False.

Answer: (A) $-4 < x$ .

Problem 8:

Solve for $x$ : $-1 – 5x \geq 14$ .

Solution:

$-1 – 5x \geq 14$

Add 1 to both sides:

$-5x \geq 15$

Divide by $-5$ :

$x \leq -3$

Answer: (D) $x \leq -3$ .

Problem 9:

Solve for $y$ : $3y + 5 \leq -10$ .

Solution:

$3y + 5 \leq -10$

Subtract 5 from both sides:

$3y \leq -15$

Divide by 3:

$y \leq -5$

Answer: (C) $y \leq -5$ .

Problem 10:

If $x^1 > 5^1$ , then:

Answer: (B) $x > 5$ .

Problem 11:

If $y > 4$ , which has the least value?

(A) $\frac{4}{y + 1}$
(B) $\frac{4}{y – 1}$
(C) $\frac{4}{y}$
(D) $\frac{y}{4}$
(E) $\frac{y + 1}{4}$

Solution:
For $y > 4$ , the smallest denominator gives the largest fraction.
Least value is $\frac{4}{y + 1}$ .

Answer: (A) $\frac{4}{y + 1}$ .

Example 1: Total Cost of Five Books

If the average cost of five books is $2.30, what is the total cost of all five books?

We use the formula for the average:

$\text{Average} = \frac{\text{Sum of items}}{\text{Number of items}}$

Substitute the known values:

$2.30 = \frac{x}{5}$

Multiply both sides by 5:

$x = 5 \times 2.30$ $x = 11.50$

Total Cost: $11.50

Example 2: Average of 0.6, 6, and 60

We use the same formula for the average:

$\text{Average} = \frac{\text{Sum of items}}{\text{Number of items}}$

Calculate the sum of the items:

$0.6 + 6 + 60 = 66.6$

Divide by the number of items (3):

$\text{Average} = \frac{66.6}{3} = 22.2$

Average: 22.2

Example III: Finding the Number of Rooms in a House

Problem:
If Jim takes a total of 20 hours to paint a house and he can paint an average room in 2.5 hours, how many rooms are in the house?

Solution:
We use the formula for the average:

\text{Average time per room} = \frac{\text{Total time}}{\text{Number of rooms}}

Substitute the known values:

2.5 = \frac{20}{\text{Number of rooms}}

Rearrange the formula to solve for the number of rooms:

\text{Number of rooms} = \frac{20}{2.5}

Perform the division:

\text{Number of rooms} = 8

Answer: There are 8 rooms in the house.

Practice Problems: Averages

What is the average of $\frac{3}{2}, \frac{5}{6},$ and $\frac{2}{3}$ ?
- Formula: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of terms}}$
- Solution: $\text{Sum} = \frac{3}{2} + \frac{5}{6} + \frac{2}{3} = \frac{9}{6} + \frac{5}{6} + \frac{4}{6} = \frac{18}{6} = 3$ $\text{Average} = \frac{3}{3} = 1$ Answer: $1$
What number must be added to $6, 16,$ and $8$ to attain an average of $13$ ?
- Formula: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of terms}}$
- Let the missing number be $x$ : $\frac{6 + 16 + 8 + x}{4} = 13$ $30 + x = 52$ $x = 22$ Answer: $22$
After taking his fourth quiz, Bill’s average dropped from $78$ to $75$ . What was Bill’s last quiz grade?
- Formula: $\text{Total Score} = \text{Average} \times \text{Number of quizzes}$
- Initial total for 3 quizzes: $78 \times 3 = 234$
- New total for 4 quizzes: $75 \times 4 = 300$
- Last quiz grade: $300 – 234 = 66$ Answer: $66$
A piece of rope $18$ feet $4$ inches long is to be cut into four equal pieces. What will be the length of each piece?
- Convert $18$ feet $4$ inches to inches: $18 \times 12 + 4 = 220 \, \text{inches}$
- Divide by 4: $\frac{220}{4} = 55 \, \text{inches}$
- Convert back to feet and inches: $55 \, \text{inches} = 4 \, \text{feet}, 7 \, \text{inches}$ Answer: $4 \, \text{feet}, 7 \, \text{inches}$
What is the average of $\frac{3}{4}, 0.64,$ and $0.87$ ?
- Convert $\frac{3}{4}$ to decimal: $0.75$
- Formula: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of terms}}$ $\text{Sum} = 0.75 + 0.64 + 0.87 = 2.26$ $\text{Average} = \frac{2.26}{3} \approx 0.753$ Answer: $0.753$
Find the average of $x, x-3, 2x-5, 2x+2,$ and $1-x$ .
- Formula: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of terms}}$ $\text{Sum} = x + (x-3) + (2x-5) + (2x+2) + (1-x) = 5x – 5$ $\text{Average} = \frac{5x-5}{5} = x-1$ Answer: $x-1$
The lowest temperatures recorded each day during a week were $-7^\circ, 5^\circ, 1^\circ, -10^\circ, -8^\circ, 2^\circ,$ and $0^\circ$ . To the nearest degree, what was the average minimum temperature?
- Formula: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of terms}}$ $\text{Sum} = -7 + 5 + 1 – 10 – 8 + 2 + 0 = -17$ $\text{Average} = \frac{-17}{7} \approx -2.43 \approx -2$ Answer: $-2^\circ$
The average of $2x-1, 4, 6, 12,$ and $13$ is $9$ . What is the value of $x$ ?
- Formula: $\text{Average} = \frac{\text{Sum of numbers}}{\text{Number of terms}}$ $\frac{(2x-1) + 4 + 6 + 12 + 13}{5} = 9$ $2x + 34 = 45$ $2x = 11$ $x = 5.5$ Answer: $5.5$
The average of $x$ and $y$ is $4$ . If $x = 5y$ , what is the value of $y$ ?
- Formula: $\frac{x+y}{2} = 4$ $\frac{5y + y}{2} = 4$ $6y = 8$ $y = \frac{4}{3}$ Answer: $\frac{4}{3}$
Three sisters weigh $108 \frac{1}{2}$ pounds, $97 \frac{1}{4}$ pounds, and $121 \frac{3}{4}$ pounds. What is their average weight?
- Convert to improper fractions: $108 \frac{1}{2} = \frac{217}{2}, \, 97 \frac{1}{4} = \frac{389}{4}, \, 121 \frac{3}{4} = \frac{487}{4}$
- Find the sum: $\frac{217}{2} + \frac{389}{4} + \frac{487}{4} = \frac{434}{4} + \frac{389}{4} + \frac{487}{4} = \frac{1310}{4}$
- Divide by 3 for the average: $\frac{1310}{4} \div 3 = \frac{1310}{12} \approx 109.17$ Answer: $109.17 \, \text{pounds}$

Practice SAT Problems

Problem 1

Question: If $\frac{4}{5} = \frac{x}{4}$ , then $x =$
(A) $\frac{5}{16}$
(B) $\frac{5}{4}$
(C) $\frac{16}{5}$
(D) $5$
(E) $16$

Solution: Cross multiply:

$4 \cdot 4 = 5 \cdot x \implies 16 = 5x \implies x = \frac{16}{5}$

Answer: (C) $\frac{16}{5}$

Problem 2

Question: Juan’s quiz scores in math were 95, 87, 84, 84, and 60. What was the average of these scores?
(A) 80.2
(B) 81
(C) 81.5
(D) 82
(E) 84

Solution: Calculate the sum of the scores:

$95 + 87 + 84 + 84 + 60 = 410$

Divide by the number of scores:

$\text{Average} = \frac{410}{5} = 82$

Answer: (D) $82$

Problem 3

Question: A woman’s bill at the food store was $15.00. She had coupons worth $3.50. If she handed the clerk $20.00, how much change did she get back?
(A) $3.50
(B) $5.00
(C) $6.50
(D) $7.50
(E) $8.50

Solution:

Total bill after coupons: $15.00 – 3.50 = 11.50$
Change from $20: $20.00 – 11.50 = 8.50$

Answer: (E) $8.50

Problem 4

Question: If $\frac{1}{3}$ of a number is 6 more than $\frac{1}{4}$ of the number, what is the number?
(A) 72
(B) 60
(C) 48
(D) 24
(E) 18

Solution: Let the number be $x$ :

$\frac{1}{3}x = \frac{1}{4}x + 6$

Multiply through by 12 to eliminate fractions:

$4x = 3x + 72 \implies x = 72$

Answer: (A) $72$

Problem 5

Question: If $x = \frac{3}{4}$ and $y = \frac{9}{8}$ , what is the value of $\frac{x}{y}$ ?
(A) 1
(B) $\frac{2}{9}$
(C) $\frac{2}{3}$
(D) $\frac{1}{3}$
(E) $\frac{27}{32}$

Solution:

$\frac{x}{y} = \frac{\frac{3}{4}}{\frac{9}{8}} = \frac{3}{4} \times \frac{8}{9} = \frac{24}{36} = \frac{2}{3}$

Answer: (C) $\frac{2}{3}$

Problem 6

Question: Evaluate $(0.53)^2 – (0.52)^2$
(A) 0.0105
(B) 0.105
(C) 1.05
(D) 10.5
(E) 105

Solution: Using the difference of squares formula $a^2 – b^2 = (a-b)(a+b)$ :

$(0.53 – 0.52)(0.53 + 0.52) = (0.01)(1.05) = 0.0105$

Answer: (A) $0.0105$

Problem 7

Question: What fraction of 8 hours is 120 seconds?
(A) $\frac{1}{60}$
(B) $\frac{1}{120}$
(C) $\frac{1}{240}$
(D) $\frac{1}{1200}$
(E) $\frac{1}{2400}$

Solution:

Convert 8 hours to seconds: $8 \times 60 \times 60 = 28,800 \, \text{seconds}$
Fraction: $\frac{120}{28,800} = \frac{1}{240}$

Answer: (C) $\frac{1}{240}$

Problem 8

Question: Sandy’s test scores average 82. If her scores are $78, 78, 92$ , what is the fourth score?
(A) 74
(B) 78
(C) 80
(D) 82
(E) 88

Solution:

Total score for 4 tests: $82 \times 4 = 328$
Sum of first three scores: $78 + 78 + 92 = 248$
Fourth score: $328 – 248 = 80$

Answer: (C) $80$

Problem 9

Question: Simplify $\frac{23}{1000} + \frac{6}{100} + \frac{7}{10}$
(A) 0.7623
(B) 0.0042
(C) 0.783
(D) 0.2367
(E) 0.327

Solution: Convert to decimals:

$\frac{23}{1000} = 0.023, \, \frac{6}{100} = 0.06, \, \frac{7}{10} = 0.7$

Add:

$0.023 + 0.06 + 0.7 = 0.783$

Answer: (C) $0.783$

Problem 10

Question: If $8x + 8y = 64$ , what is the average of $x$ and $y$ ?
(A) 16
(B) 4
(C) 8
(D) 2
(E) Cannot be determined

Solution: Simplify:

$x + y = \frac{64}{8} = 8$

Average:

$\frac{x + y}{2} = \frac{8}{2} = 4$

Answer: (B) $4$

Problem 11

Question: If 8 and 36 each divide $N$ without remainder, what is $N$ ?
(A) 64
(B) 72
(C) 108
(D) 288
(E) Cannot be determined

Solution: The least common multiple of 8 and 36 is:

$\text{LCM} = 2^3 \cdot 3^2 = 72$

Answer: (B) $72$

Problem 12

Question: What is the thickness of 3 sheets of paper if 250 sheets are 2.0 inches thick?
(A) 0.008
(B) 0.024
(C) 0.04
(D) 0.4
(E) 0.8

Solution:

Thickness of 1 sheet: $\frac{2.0}{250} = 0.008 \, \text{inches}$
Thickness of 3 sheets: $3 \times 0.008 = 0.024$

Answer: (B) $0.024$

Problem 13

Question: If the average of $5, 6, 8, x, 12$ is 12, find $x$ .
(A) 17
(B) 25
(C) 29
(D) 39
(E) 60

Solution:

Total sum: $\text{Sum} = 12 \times 5 = 60$
Known sum: $5 + 6 + 8 + 12 = 31$
Solve for $x$ : $x = 60 – 31 = 29$

Answer: (C) $29$

Basic Algebra Concepts: Lesson 1

Basic Algebra Concepts: Lesson 1

Core Mathematical Operations

Key Algebraic Rules

Example: Solving a Polynomial

Laws of Exponents

Square Roots (Radicals)

Place Value

Translating Words and Phrases into Algebraic Symbols

Examples with Completed Calculations

Examples with Alternative Formatting

Example 1: Determining Tank Capacity

Example 2: Consecutive Odd Numbers

Example III: Dividing Money Among Michael, Paul, and Susan

Example IV: Age Puzzle

For problems 6-20, enter your solutions into the grid provided below each question.

Problems 6–20: Enter your solutions into the grid provided below each question.

Understanding Inequalities and Their Manipulations

Rules for Solving Inequalities

1. Adding or Subtracting a Number

2. Multiplication or Division by a Positive Number

3. Multiplication or Division by a Negative Number

4. Taking the Square Roots

5. Squaring Both Sides

6. Inverting Both Sides

7. Absolute Values

Additional Inequality Properties

Example: Comparing Values with Inequalities

Problem:

Solution:

Special Inequality Symbols

Examples with ≤\leq≤ and ≥\geq≥

Example 1: Combining Inequalities

Example 2: Contradictory Inequalities

Example 3: True and False Comparisons

Example Problems: Inequalities and Logical Reasoning

Problem 1: Solve for xx

Problem 2: Logical Reasoning

Important Note:

Practice Problems with Solutions and Explanations:

Problem 1:

Problem 2:

Problem 3:

Problem 4:

Problem 5:

Problem 6:

Problem 7:

Problem 8:

Problem 9:

Problem 10:

Problem 11:

Example 1: Total Cost of Five Books

Example 2: Average of 0.6, 6, and 60

Example III: Finding the Number of Rooms in a House

Practice Problems: Averages

Practice SAT Problems

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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