**Algebraic Expressions**

A mathematical expression that contains a variable is called an algebraic expression. Some examples of algebraic expressions are x^{2} , 3x Â–â€“ 2y, 2z$({y}^{3}-\frac{1}{{Z}^{2}})$. Two algebraic expressions are called like terms if both the variable parts and the exponents are identical. That is, the only parts of the expressions that can differ are the coefficients. For example, 5y^{3} and $\frac{3}{2}{y}^{3}$ are like terms, as are x + y^{2} and -7(x + y^{2}). However, x^{3} and y^{3} are not like terms, nor are x Â–â€“ y and 2 Â–â€“ y.

**Adding and Subtracting Algebraic Expressions**

Only like terms may be added or subtracted. To add or subtract like terms, merely add or subtract their coefficients:

x^{2} + 3x^{2} = (1 + 3)x^{2} = 4x^{2}

$2\sqrt{x}-5\sqrt{x}=\left(2-5\right)\sqrt{x}=-3\sqrt{x}$

$.5{\left(x+\frac{1}{y}\right)}^{2}+.2{\left(x+\frac{1}{y}\right)}^{2}=\left(.5+.2\right){\left(x+\frac{1}{y}\right)}^{2}=.7{\left(x+\frac{1}{y}\right)}^{2}$

(3x^{3} + 7x^{2} + 2x + 4) + (2x^{2} - 2x - 6) = 3x^{3} + (7 + 2)x^{2} + (2 - 2)x + (4 - 6) = 3x^{3} + 9x^{2} - 2

You may add or multiply algebraic expressions in any order. This is called the **commutative property**:

**Caution:**the commutative property does not apply to division or subtraction: 2 = 6 Ã· 3 â‰ 3 Ã· 6 = 1 2

and Â–â€“1 = 2 Â–â€“ 3 â‰ 3 Â–â€“ 2 = 1.

When adding or multiplying algebraic expressions, you may regroup the terms. This is called the **associative property**:

Notice in these formulas that the variables have not been moved, only the way they are grouped has changed: on the left side of the formulas the last two variables are grouped together, and on the right side of the formulas the first two variables are grouped together.

For example, (x Â–â€“ 2x) + 5x = (x + [Â–â€“2x]) + 5x = x + (Â–â€“2x + 5x) = x + 3x = 4x

and

2(12x ) = (2 Â· 2)x = 24x

**The associative property doesn't apply to division or subtraction:**

4 = 8 Ã· 2 = 8 Ã· (4 Ã· 2) â‰ (8 Ã· 4)

and

Â–â€“6 = Â–â€“3 Â–â€“ 3 = (Â–â€“1 Â–â€“ 2) Â–â€“ 3 â‰ Â–â€“1 Â–â€“ (2 Â–â€“ 3) = Â–â€“1 Â–â€“ (Â–â€“1) = Â–â€“1 + 1 = 0.

Notice in the first example that we changed the subtraction into negative addition: (x Â–â€“ 2x) = (x + [Â–â€“ 2x]). This allowed us to apply the associative property over addition.

**Parentheses**

5x + (y Â–â€“ (2x Â–â€“ 3x)) = 5x + (y Â–â€“ (Â–â€“x)) = 5x + (y + x) = 6x + y

Sometimes when an expression involves several pairs of parentheses, one or more pairs are written as brackets. This makes the expression easier to read:

2x(x Â–â€“ [y + 2(x Â–â€“ y)]) =

2x(x Â–â€“ [y + 2x Â–â€“ 2y]) =

2x(x Â–â€“ [2x Â–â€“ y]) =

2x(x Â–â€“ 2x + y) =

2x(Â–â€“x + y) =

-2x

^{2}+ 2xy

**Order of Operations: (PEMDAS)**

**PEMDAS**

**P**lease

**E**xcuse

**M**y

**D**ear

**A**unt

**S**ally

This mnemonic isnÂ’â€™t quite precise enough. Multiplication and division are actually tied in order of operation, as is the pair addition and subtraction. When multiplication and division, or addition and subtraction, appear at the same level in an expression, perform the operations from left to right. For example, 6 Ã· 2 Ã— 4 = (6 Ã· 2) Ã— 4 = 3 Ã— 4 = 12. To emphasize this left-to-right order, we can use parentheses in the mnemonic:

**PE**(

**MD**)(

**AS**).

**Example**:

2 - (5 - 3^{3}[4 Ã· 2 + 1]) = ?

(A) Â–â€“21

(B) 32

(C) 45

(D) 60

(E) 78

2 - (5 - 3^{3}[4 Ã· 2 + 1]) =

By performing the division within the innermost parentheses

2 - (5 - 3^{3}[2 + 1]) =

By performing the addition within the innermost parentheses

2 - (5 - 3^{3}[3]) =

By performing the exponentiation

2 Â–â€“ (5 Â–â€“ 27[3]) =

By performing the multiplication within the parentheses

2 Â–â€“ (5 Â–â€“ 81) =

By performing the subtraction within the parentheses

2 Â–â€“ (Â–â€“76) =

By multiplying the two negatives

2 + 76 =

78

The answer is (E).

**Foil Multiplication**

**F**irst,

**O**uter,

**I**nner,

**L**ast

Simplifying the right side yields (x + y)(x + y) = x

^{2}+ 2xy + y

^{2}. For the product (x - y)(x - y) we get (x - y)(x - y) = x

^{2}- 2xy + y

^{2}. These types of products occur often, so it is worthwhile to memorize the formulas. Nevertheless, you should still learn the FOIL method of multiplying because the formulas do not apply in all cases.

**Examples (FOIL):**

(2 - y)(x - y^{2}) = 2x - 2y^{2} - xy + yy^{2} = 2x - 2y^{2} - xy +y^{3}

$\left(\frac{1}{x}-y\right)\left(x-\frac{1}{y}\right)=\frac{1}{x}x-\frac{1}{x}\frac{1}{y}-xy+y\frac{1}{y}=1-\frac{1}{xy}-xy+1=2-\frac{1}{xy}-xy$

${\left(\frac{1}{2}-y\right)}^{2}=\left(\frac{1}{2}-y\right)\left(\frac{1}{2}-y\right)={\left(\frac{1}{2}\right)}^{2}-2\left(\frac{1}{2}\right)y+{y}^{2}=\frac{1}{4}-y+{y}^{2}$

**Division of Algebraic Expressions**

$\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}$

This formula generalizes to any number of terms.

**Example 1:**

$\frac{{x}^{2}+y}{x}=\frac{{x}^{2}}{x}+\frac{y}{x}={x}^{2-1}+\frac{y}{x}=x+\frac{y}{x}$

$\frac{{x}^{2}+2y-{x}^{3}}{{x}^{2}}=\frac{{x}^{2}}{{x}^{2}}+\frac{2y}{{x}^{2}}-\frac{{x}^{3}}{{x}^{2}}={x}^{2-2}+\frac{2y}{{x}^{2}}-{x}^{3-2}={x}^{0}+\frac{2y}{{x}^{2}}-x=1+\frac{2y}{{x}^{2}}-x$

When there is more than a single variable in the denomination, we usually factor the expression and then cancel, instead of using the above formula.

**Example 2:**

$\frac{{x}^{2}-2x+1}{x-1}=$

(A) x + 1

(B) Â–â€“x Â–â€“ 1

(C) Â–â€“x + 1

(D) x Â–â€“ 1

(E) x Â–â€“ 2

$\frac{{x}^{2}-2x+1}{x-1}=\frac{\left(x-1\right)\left(x-1\right)}{x-1}=x-1$. The answer is (D).