Evaluating Algebraic Expressions

In the last section, we simplified expressions using the order of operations. In this section, we’ll evaluate expressions—again following the order of operations.

To evaluate an algebraic expression means to find the value of the expression when the is replaced by a given number. To evaluate an , we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.

Example

Evaluate \(x+7\) when

\(\phantom{\rule{0.2em}{0ex}}x=3\)
\(\phantom{\rule{0.2em}{0ex}}x=12\)

Solution

To evaluate, substitute \(3\) for \(x\) in the expression, and then simplify.


Substitute.
Add.

When \(x=3,\) the expression \(x+7\) has a value of \(10.\)

To evaluate, substitute \(12\) for \(x\) in the expression, and then simplify.


Substitute.
Add.

When \(x=12,\) the expression \(x+7\) has a value of \(19.\)

Notice that we got different results for parts and even though we started with the same expression. This is because the values used for \(x\) were different. When we evaluate an expression, the value varies depending on the value used for the variable.

Example

Evaluate \(9x-2,\text{when}\phantom{\rule{0.2em}{0ex}}\)

\(\phantom{\rule{0.2em}{0ex}}x=5\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}\)
\(\phantom{\rule{0.2em}{0ex}}x=1\)

Solution

Remember \(ab\) means \(a\) times \(b,\) so \(9x\) means \(9\) times \(x.\)

To evaluate the expression when \(x=5,\) we substitute \(5\) for \(x,\) and then simplify.



Multiply.
Subtract.

To evaluate the expression when \(x=1,\) we substitute \(1\) for \(x,\) and then simplify.



Multiply.
Subtract.

Notice that in part that we wrote \(9\cdot 5\) and in part we wrote \(9\left(1\right).\) Both the dot and the parentheses tell us to multiply.

Example

Evaluate \({x}^{2}\) when \(x=10.\)

Solution

We substitute \(10\) for \(x,\) and then simplify the expression.



Use the definition of exponent.
Multiply.

When \(x=10,\) the expression \({x}^{2}\) has a value of \(100.\)

Example

\(\text{Evaluate}\phantom{\rule{0.2em}{0ex}}{2}^{x}\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}x=5.\)

Solution

In this expression, the variable is an exponent.



Use the definition of exponent.
Multiply.

When \(x=5,\) the expression \({2}^{x}\) has a value of \(32.\)

Example

\(\text{Evaluate}\phantom{\rule{0.2em}{0ex}}3x+4y-6\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}x=10\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=2.\)

Solution

This expression contains two variables, so we must make two substitutions.



Multiply.
Add and subtract left to right.

When \(x=10\) and \(y=2,\) the expression \(3x+4y-6\) has a value of \(32.\)

Example

\(\text{Evaluate}\phantom{\rule{0.2em}{0ex}}2{x}^{2}+3x+8\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}x=4.\)

Solution

We need to be careful when an expression has a variable with an exponent. In this expression, \(2{x}^{2}\) means \(2\cdot x\cdot x\) and is different from the expression \({\left(2x\right)}^{2},\) which means \(2x\cdot 2x.\)



Simplify \({4}^{2}\).
Multiply.
Add.

Evaluating Algebraic Expressions