Expression
An algebraic expression is a combination of variables, numbers, arithmetic operations, powers, roots, and more. Within algebra, there are many forms of expressions that may have specific names, such as polynomials. Below are a few examples of algebraic expressions:
x + 7
x2 + 3x + 4
Evaluating expressions
Expressions can be part of equations, or they can also be evaluated for a given value. This means to substitute the given value into the expression then perform the operations in the expression.
Examples
1. Evaluate the following expression given that x = 5
(2x + 7) × x
(2(5) + 7) × 5 = (17) × 5 = 85
Another way that this problem could have been posed is using the following notation:
[(2x + 7) × x] | x = 5
The vertical bar means to evaluate the expression given what comes after the bar.
2. (2x2 + 7x) | x = 5
2(5)2 + 7(5) = 50 + 35 = 85
From the above examples we can see that:
(2x + 7) × x = 2x2 + 7x
They are equivalent expressions.
Equivalent expressions
Equivalent expressions are expressions in which for some given variable (or variables), such as x, the values of the expressions are equal.
2x + 2 = 2(x + 1)
For any value of x, the expressions above have the same value; they are just in different forms. We can test this by plugging a few values in for x.
If x = 1: 2(1) + 2 = 2(1 + 1)
4 = 4
If x = -5: 2(-5) + 2 = 2(-5 + 1)
-8 = -8
We can do this for many more numbers, but the result will be the same.
Writing expressions in different but equivalent forms can be helpful in certain mathematical contexts, including solving equations in algebra. Having a factored, equivalent form of an expression is one commonly used method for solving algebraic equations involving variables raised to a power.
Example
Solve 2x + 2 = (x + 1)(x + 7)
Expand the right side of the equation, then group the like terms:
2x + 2 = x2 + 8x + 7
x2 + 6x + 5 = 0
Factor the left side of the equation to find an equivalent expression:
x2 + 6x + 5 = (x + 1)(x + 5) = 0
Now that the equation is factored, we can solve for x by dividing each term and setting the remainding term equal to 0. Note that we must do this for each term or we would lose one of the solutions to the problem.
x + 1 = 0
x = -1
and
x + 5 = 0
x = -5
Therefore, the solutions to the above equation are x = -1 and x = -5