# 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 $e=5$.