foundations recap

Foundations Recap

Algebra becomes less difficult because of one complicated rule, and more because a challenging problem often combines several simple rules at once. A forgotten exponent law, a misplaced negative sign, an incorrect distribution, or a small mistake with fractions can turn a manageable problem into a difficult one.

This Foundations Recap is designed to prevent that.

Think of it as your algebra starter pack: a compact but comprehensive review of the essential algebraic ideas built throughout your school years. Instead of repeatedly going back to basic concepts whenever a difficult SAT or GRE question exposes a gap, use this section to refresh the underlying rules, patterns, and techniques first.

That is why strong fundamentals matter.

Think of it as your algebra toolbox. The stronger these tools become, the more attention you can devote to the actual problem rather than the mechanics of solving it.

1. Algebraic Expressions

An algebraic expression represents a quantity using numbers, variables, and mathematical operations.

Example:

3x + 5

could represent a price that consists of a fixed $5 fee plus $3 for every unit purchased.

Expression vs Equation: An expression does not contain an equals sign. Once we write:

3x + 5 = 20

we have an equation. As a result:

• Expression: 3x + 5
• Equation: 3x + 5 = 20

Key terms. In:

4x² − 7x + 9

• x is the variable.
• 4 and −7 are coefficients.
• 9 is a constant.
• 4x², −7x, and 9 are terms.

Like terms. Only terms with exactly the same variable structure can be combined.

3x + 5x = 8x

but:

3x + 5x²

cannot be combined.

Example (gym fee). The following example explains how terms can be combined into an equation to find the missing element. Suppose a gym charges a $15 registration fee and $20 per month. If a customer stays for m months, the total cost is:

C = 20m + 15

For 6 months:

C = 20(6) + 15

C = 120 + 15

C = 135

So the six-month cost is $135. The important idea is that algebra allows us to create a formula that works for any number of months, not just six.

Common trap: Do not combine unlike terms:

4x + 3 ≠ 7x

The 3 has no x, so these are not like terms.

A simple formula that is used to solve several algebraic expressions:

(a + b)(a − b) = a² − b²

However, students shouldn’t mix it with:

(a − b)² = a² − 2ab + b²

So stating (a + b)(a − b) = a² − b² = (a − b)² is inaccurate.

Example. To solve (x + 6)(x − 6)(x + 1)(x − 1), we must use the aforementioned rule (a + b)(a − b) = a² − b². Therefore,

(x + 6)(x − 6) = (x² − 36)

(x + 1)(x − 1) = (x² − 1)

Now multiply both the resulting brackets:

x⁴ − x² − 36x² + 36

Adding x² and −36x²:

x⁴ − 37x² + 36

Example (factorizing). Similarly, if the question asks: Factorize x² − 25. We might reverse the process of the earlier question:

(x)² − (5)²

As a² − b² = (a + b)(a − b), therefore:

(x + 5)(x − 5)

2. Order of Operations

When an expression contains several operations, they must be performed in the correct order. Use:

1. Parentheses/grouping
2. Exponents
3. Multiplication and division, left to right
4. Addition and subtraction, left to right

Example (discount). Suppose a $80 product receives a 20% discount and then a $5 delivery charge is added. The mathematical expression is:

80(1 − 0.20) + 5

First calculate inside the parentheses:

80(0.80) + 5

Then multiply:

64 + 5 = 69

The final price is $69.

Important point: Multiplication does not always come before division. For:

24 ÷ 6 × 2

work from left to right:

4 × 2 = 8

Be careful with negative signs and exponents:

−3² = −9

but:

(−3)² = 9

The parentheses determine whether the negative sign is included in the base.

3. Distributive Property

The distributive property allows a number or expression outside parentheses to multiply every term inside.

a(b + c) = ab + ac

Example (boxes). Suppose a company buys x boxes of a product. Each box costs $25, but each box also has a $3 handling charge. The cost per box is:

25 + 3

For x boxes:

x(25 + 3)

Using distribution:

25x + 3x = 28x

So the total cost is 28x.

Negative distribution. Consider:

−2(x − 4)

The −2 applies to both terms:

−2x + 8

Common trap:

• Incorrect: −(x − 5) = −x − 5
• Correct: −(x − 5) = −x + 5

4. Combining Like Terms

Like terms represent the same type of quantity and can be combined.

4x + 7x = 11x

Example (warehouse). Suppose a warehouse has 4x boxes in one section, 7x boxes in another section, and 12 loose boxes. Total:

4x + 7x + 12 = 11x + 12

You can combine the two groups of x-boxes, but the 12 loose boxes remain separate.

With multiple variables:

3x + 4y − 2x + 7y

becomes:

x + 11y

Common trap: x² + x cannot become 2x². The terms represent different powers of x.

5. Exponent Rules

Exponents tell us how many times a quantity is multiplied by itself.

Product rule

a^m · aⁿ = a^m+n

Quotient rule

a^maⁿ = a^m−n

Power of a power

(a^m)ⁿ = a^mn

Zero exponent

a⁰ = 1 (for a ≠ 0)

Negative exponent

a⁻ⁿ = 1aⁿ

Example (doubling). Suppose a quantity doubles every year. Starting with P, after one year:

2P

After two years:

2(2P) = 2²P

After three years:

2³P

After n years:

P(2ⁿ)

The exponent therefore represents the number of repeated growth periods.

Common trap: Exponent rules do not apply to addition:

x² + x³ ≠ x⁵

But:

x² · x³ = x⁵

6. Algebraic Fractions

Fractions behave the same way in algebra as they do with numbers, but variables may appear in the numerator or denominator.

Example (sharing a cost). Suppose a $600 expense is divided equally among x people. Each person’s share is:

600x

If 10 people share the cost:

60010 = 60

So each person pays $60.

Adding algebraic fractions. Suppose:

x4 + x6

The common denominator is 12:

3x12 + 2x12 = 5x12

Dividing fractions

ab ÷ cd = ab × dc

Critical restriction: A denominator cannot equal zero. For:

5x − 3

we must have x ≠ 3.

Common trap: Do not cancel terms across addition. For example:

x + 3x

cannot be simplified by “canceling x.” Cancellation applies to factors, not individual terms being added.

7. Factoring

Factoring rewrites an expression as a product of simpler expressions. It is essentially the reverse of distribution.

6x + 12 = 6(x + 2)

Example (garden). Suppose a rectangular garden has area:

x² + 5x + 6

We want to express its dimensions as two algebraic factors. Find two numbers whose product is 6 and whose sum is 5. Those numbers are 2 and 3. Therefore:

x² + 5x + 6 = (x + 2)(x + 3)

So the rectangle could have dimensions x + 2 and x + 3. This is one reason factoring is so useful in geometry, algebra, and quadratics.

Another example. How many distinct factors are there for the following expression:

(x² − 4)(x² + 4x + 4)

As we can notice, the second part of the expression can be transformed into a perfect-square factor:

x² + 4x + 4 = (x + 2)(x + 2)

As a result, the expression becomes:

(x² − 4)(x + 2)(x + 2)

Now, we can factor the first part using the difference of squares rule:

x² − 4 = (x − 2)(x + 2)

Therefore, the expression becomes:

(x − 2)(x + 2)(x + 2)(x + 2)

The distinct factors are x − 2 and x + 2. Although the factor x + 2 appears three times, it is counted only once because the question asks for distinct factors.

Final answer: 2 distinct factors.

Difference of squares

a² − b² = (a − b)(a + b)

For example, x² − 25 becomes:

(x − 5)(x + 5)

Common trap: Do not forget to check for a common factor before attempting more complicated factoring.

8. Substitution

Substitution means replacing a variable with a known value or expression.

Example (trip cost). Suppose the cost of a trip is:

C = 50 + 12d

where d is the number of days. For a 5-day trip:

C = 50 + 12(5)

C = 50 + 60

C = 110

So the cost is $110.

Substituting a negative number. If x = −4, then:

x² = (−4)² = 16

Always use parentheses when replacing a variable with a negative number.

Substituting an expression. If x = y + 2, then:

x² = (y + 2)²

and:

x² = y² + 4y + 4

Substitution often lets you turn a complicated expression into something simpler without solving for every variable.

9. Absolute Value

Absolute value represents distance from zero. Therefore:

|5| = 5 and |−5| = 5

Therefore:

|x| ≥ 0

This idea of distance is the key to understanding nearly every absolute-value problem.

Absolute value as distance. Think of |x| as asking: How far is x from zero? For example:

|−8| = 8

because −8 is 8 units away from zero. Similarly:

|3| = 3

because 3 is 3 units away from zero. This also explains why opposite numbers have the same absolute value:

|7| = |−7| = 7

Absolute value of an expression. Absolute value does not have to be applied to a single number. For example, |x − 5| represents the distance between x and 5. Therefore:

|x − 5| = 3

means that x is 3 units away from 5. There are two possibilities:

x = 8 or x = 2

because both 8 and 2 are exactly 3 units from 5.

Absolute-value equations. The general form |x| = a, where a > 0, has two solutions:

x = a or x = −a

More generally, |x − a| = b means that x is b units away from a. Therefore:

x − a = b or x − a = −b

which gives:

x = a + b or x = a − b

Example (temperature). Suppose a machine operates at a target temperature of 20°C. If the temperature is exactly 4°C away from the target, we can write:

|T − 20| = 4

This gives two possibilities:

T − 20 = 4, so T = 24

T − 20 = −4, so T = 16

Thus, the two possible temperatures are T = 16°C or T = 24°C. The absolute value tells us the distance from the target temperature, not whether the temperature is above or below it.

Absolute-value inequalities. Absolute value becomes especially important when describing a range of values. Consider:

|x| < 5

This means that the distance of x from zero is less than 5. Therefore:

−5 < x < 5

For example, x = 3 satisfies this inequality because 3 is less than 5 units from zero.

Example (opposite temperatures). Suppose the temperature is either 6°C or −6°C. Both temperatures are 6° away from zero. Therefore:

|6| = |−6| = 6

Common trap: Absolute value does not mean “remove the negative sign” in every algebraic situation. It represents distance and therefore can produce multiple solutions.

10. Algebraic Manipulation

Algebraic manipulation means transforming an expression or equation into another equivalent form. The goal is often not to “solve for x” but to transform the information into a form that reveals what the question asks.

Example. Find the simplest form of:

(x + 1)²(x² − x)x²(x² − 1)

Solution. Factor out common terms from the numerator:

x² − x = x(x − 1)

Factor the difference of squares in the denominator:

x² − 1 = (x − 1)(x + 1)

Substitute these factored forms into the expression:

(x + 1)² · x(x − 1)x²(x − 1)(x + 1)

Group common factors in the numerator and denominator:

x(x + 1)²(x − 1)x²(x + 1)(x − 1)

Cancel out the common factors (x − 1), (x + 1), and x:

x + 1x or 1 + 1x

11. Translating Words into Algebra

Many standardized-test questions hide algebra inside English. A lot of times students are given a problem in words and a solution is expected out of it. The first task is often to translate the wording into a mathematical relationship.

Example (mobile plan). A mobile company charges $30 per month plus $5 for every additional gigabyte of data. If g is the number of additional gigabytes, the monthly bill is:

B = 30 + 5g

If the bill is $55:

55 = 30 + 5g

25 = 5g

g = 5

So the customer used 5 additional gigabytes.

12. Equivalent Expressions

Two expressions are equivalent if they produce the same value for every allowed value of the variable. For example, 2(x + 3) and 2x + 6 are equivalent.

Example (rectangle). Suppose a rectangle has dimensions x + 3 and x + 5. Its area is:

(x + 3)(x + 5)

Expanding:

x² + 8x + 15

Therefore (x + 3)(x + 5) and x² + 8x + 15 are equivalent expressions. One form may be better for understanding the dimensions, while the other may be better for calculation.

13. Essential Algebraic Identities

Certain patterns occur so frequently that recognizing them should become automatic.

Difference of squares

a² − b² = (a − b)(a + b)

Example: x² − 49 = (x − 7)(x + 7)

Square of a sum

(a + b)² = a² + 2ab + b²

Square of a difference

(a − b)² = a² − 2ab + b²

Example (101 squared). Suppose you need 101². Instead of multiplying 101 × 101, write 101 = 100 + 1. Therefore:

101² = (100 + 1)²

= 100² + 2(100)(1) + 1²

= 10000 + 200 + 1

= 10201

Recognizing algebraic structure can make calculations much faster.

14. Solving Simple Equations

Even though linear equations have their own major section, the foundation recap should establish the basic principle. An equation states that two expressions have equal values.

Example 1. Solve:

x + 7 = 15

Subtract 7 from both sides:

x = 8

Example (cinema). Suppose a cinema sells tickets for $12 each and collects $60 in total. If n tickets were sold:

12n = 60

Divide both sides by 12:

n = 5

Five tickets were sold.

Example 2. Factorize:

x² − 16x + 64

We might use a concept here known as the “middle term breaking”. This method relies on finding two numbers whose product equals the constant term (the third or last one) and whose sum equals the middle coefficient.

For instance, take the factors of the last term here, 64. As we can see, the number 64 has two equal factors that add up to the middle coefficient, 16. To obtain a sum of −16 and a product of 64, we use −8 and −8, which add up to −16 when combined and multiply to 64.

x² − 8x − 8x + 64

x(x − 8) − 8(x − 8)

(x − 8)(x − 8)

we get:

= (x − 8)²

The balance principle: An equation behaves like a balanced scale. If you add 5 to one side, you must add 5 to the other. If you multiply one side by 3, you must multiply the other side by 3. This principle remains true regardless of how complicated the equation becomes.

15. Ratios and Proportional Algebra

Ratios connect naturally with algebra. If two quantities are in the ratio 3 : 5, they can be represented as:

3k and 5k

for some common multiplier k.

Example (class). Suppose the ratio of boys to girls in a class is 3 : 5. If there are 24 students in total:

3k + 5k = 24

8k = 24

k = 3

Therefore 3k = 9 boys and 5k = 15 girls.

Go through the detailed version of ratios and proportion. Click here.

16. Algebraic Structure and Pattern Recognition

This is perhaps the most important foundation skill for SAT and GRE. Before calculating, look for recognizable structure.

Example 1.

49x² − 81

Recognize:

(7x)² − 9²

Therefore:

(7x − 9)(7x + 9)

Example 2.

x² + 10x + 25

Recognize:

x² + 2(5)x + 5²

Therefore:

(x + 5)²

Example 3.

12x² + 18x

Before doing anything complicated, identify the greatest common factor:

6x(2x + 3)