Algebraic Expressions

An algebraic expression is a mathematical phrase that combines numbers, variables (represented by letters), and operation symbols such as $ + $, $ − $, $ · $, $ : $, along with round, square, or curly brackets.

It's used to describe relationships between quantities that can vary.

Here’s a straightforward example of an algebraic expression:

$$ 2x^2 + xy - x $$

In algebra, the multiplication symbol is often omitted. So $ xy $ is understood to mean $ x \cdot y $.

Note. When you substitute numbers for the variables, the algebraic expression becomes a numerical one. For instance, if $ x = 3 $ and $ y = 4 $, then: $$ 2x^2 + xy - x = 2(3)^2 + 3 \cdot 4 - 3 = 18 + 12 - 3 = 27 $$

Types of Algebraic Expressions

There are several categories of algebraic expressions:

• Rational expressions
  These involve only the basic algebraic operations: addition, subtraction, multiplication, division, and integer exponents. For example, $ x + y $ is rational because it’s a simple sum of variables. Similarly, $ \frac{x+1}{y-2} $ and $ (x + 2y)^3 $ are both rational expressions.
• Irrational expressions
  These include roots or radicals with variables inside. They tend to be more complex to work with. For example, $ \sqrt{x+2} $ is irrational because it contains a square root involving a variable.
• Polynomial (or integral) expressions
  These do not contain variables in the denominator or variables raised to negative exponents. Examples include $ x^2 + y^2 $ and $ (2x + 1)^2 $. Even if a number appears in the denominator, as in $ \frac{x + y}{2} $, the expression is still considered a polynomial because the denominator contains no variables.
• Fractional expressions
  These are expressions where variables appear in the denominator or are raised to negative exponents. For example, $ \frac{x+1}{x-1} $ is fractional because the variable $ x $ appears in the denominator. Likewise, $ 2y x^{-1} $ is fractional because the negative exponent means $ \frac{2y}{x} $.

  Note. Fractional expressions are not always defined. If the denominator becomes zero, the expression is undefined due to division by zero. For example, $ \frac{x+1}{x-1} $ is undefined when $ x = 1 $, because the denominator $ x - 1 $ becomes zero. To prevent this, we set a condition for existence: $ x \ne 1 $. It’s essential to examine expressions carefully to ensure denominators remain non-zero and variable values are valid.

A Worked Example

Consider the algebraic expression:

\[ E(x, y) = \frac{2x^2 - 3xy + y^2}{x - y} + \frac{x^2 - y^2}{x + y} \]

This is a rational fractional expression. It's rational because it contains no roots, and fractional because variables appear in the denominators.

Since it's a fractional expression, we must establish conditions for existence to avoid division by zero. That is, the denominators must not be equal to zero:

\( x - y \neq 0 \), so \( x \neq y \)

\( x + y \neq 0 \), so \( x \neq -y \)

Now, we can simplify the expression step by step.

We begin by factoring the numerators where possible. Here, \( x^2 - y^2 \) is a difference of squares, so:
\( x^2 - y^2 = (x - y)(x + y) \)

\[ E(x, y) = \frac{2x^2 - 3xy + y^2}{x - y} + \frac{(x - y)(x + y)}{x + y} \]

This allows us to simplify the second fraction:

\[ E(x, y) = \frac{2x^2 - 3xy + y^2}{x - y} + (x - y) \]

Now we combine the terms over a common denominator:

\[ E(x, y) = \frac{2x^2 - 3xy + y^2 + (x - y)^2}{x - y} \]

We expand the squared binomial: $ (x - y)^2 = x^2 - 2xy + y^2 $

\[ E(x, y) = \frac{2x^2 - 3xy + y^2 + x^2 - 2xy + y^2}{x - y} \]

Combining like terms in the numerator:

\[ E(x, y) = \frac{3x^2 - 5xy + 2y^2}{x - y} \]

This is the simplified form of the expression.

At this point, we can evaluate the expression using values that satisfy the existence conditions. For instance, $ x = 3 $ and $ y = 1 $ are acceptable because $ x \ne y $ and $ x \ne -y $. Substituting these into the expression gives: \[ E(3, 1) = \frac{3 \cdot 3^2 - 5 \cdot 3 \cdot 1 + 2 \cdot 1^2}{3 - 1} \] \[ E(3, 1) = \frac{3 \cdot 9 - 15 + 2}{2} \] \[ E(3, 1) = \frac{27 - 15 + 2}{2} \] \[ E(3, 1) = \frac{14}{2} = 7 \] So, the final value when $ x = 3 $ and $ y = 1 $ is: $$ E(3, 1) = 7 $$

And so on.