Evaluating expressions with one variable

CCSS Math: 6.EE.A.2c
A mixture of explanations, examples, and practice problems to have you evaluating expressions with one variable in no time!

How to evaluate an expression with one variable

Let's say we want to evaluate the expression $a + 4$. Well, first we need to know the value of the variable $a$. For example, to evaluate the expression when $\blueD {a = 1}$, we just replace $\blueD a$ with $\blueD 1$:
\begin{aligned} &\blueD a + 4 \\\\ =&\blueD1 + 4~~~~~~~~\gray{\text{Replace }\blueD{a} \text{ with } \blueD{1}\text{.}} \\\\ =&5 \end{aligned}
So, the expression $a + 4$ equals $5$ when $a = 1$.
We can just as easily evaluate $a + 4$ when $\blueD {a = 5}$:
\begin{aligned} &\blueD a + 4 \\\\ =&\blueD5 + 4~~~~~~~~\gray{\text{Replace }\blueD{a} \text{ with } \blueD{5}\text{.}} \\\\ =&9 \end{aligned}
So, the expression $a + 4$ equals $9$ when $a = 5$.

Evaluating an expression with multiplication

You might be asked to "Evaluate $3x$ when $x = 5$."
Notice how the number $3$ is right next to the variable $x$ in the expression $3x$. This means "$3$ times $x$". The reason we do this is because the old way of showing multiplication with the symbol $\times$ looks confusingly similar to the variable $x$.
Okay, so now let's solve the problem:
\begin{aligned} &3\blueD x \\\\ =& 3 \cdot \blueD5~~~~~~~~\text{Replace }\blueD{x} \text{ with } \blueD{5}\text{.} \\\\ =&15 \end{aligned}
So, the expression $3x$ equals $15$ when $x = 5$.

New ways to show multiplication

Hold on a second! Did you notice that we wrote "$3$ times $\blueD 5$" as $3 \cdot \blueD 5$ instead of as $3 \times \blueD 5$? Using a dot instead of the symbol $\times$ is another new way of showing multiplication:
$3 \cdot \blueD 5 = 15$
Parentheses can also be used to show multiplication:
$3(\blueD 5) = 15$
Let's summarize the new ways of showing multiplication that we learned.
Old wayNew way
With a variable$3 \times x$$3x$
Without variable$3 \times 5$$3\cdot 5$ or $3(5)$

Evaluating expressions where order of operations matter

For more complex expressions, we'll have to be sure to pay close attention to order of operations. Let's take a look at an example:
Evaluate $5 + 3e$ when $\blueD{e=4}$.
\begin{aligned} &5+3\blueD e \\\\ =&5 + 3 \cdot \blueD 4~~~~~~~~\gray{\text{Replace }\blueD{e} \text{ with } \blueD{4}\text{.}} \\\\ =&5 + 12 ~~~~~~~~\text{\gray{Multiply first (order of operations)}} \\\\ =&17 \end{aligned}
So, the expression $5 + 3e$ equals $17$ when $e = 4$.
Notice how we had to be careful to think about order of operations when evaluating. A common wrong answer is $\redD{32}$, which comes from first adding $5$ and $3$ to get $8$ then multiplying $8$ by $4$ to get $\redD{32}$.

Let's practice!

Problem 1
Evaluate the expression $9 - z$ when $z = 4$.

Challenge problems

Challenge problem 1
Evaluate $e \cdot e - 5e$ when $e=5$.