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Section 5.1 This is a That

We have previously seen how we can substitute a number for a variable. This led us to write sentences like "If \(x = 3\text{,}\) then \(2x + 1 = 7\text{.}\)"

We are not restricted to just single variable expressions. It is often the case that a problem is modeled using multiple variables. Algebraically, this is just a matter of making sure you use the correct value for the correct variable. But there are some potential pitfalls as the substitutions become more complex.

Activity 5.1.1. Substituting for Multiple Variables.

The problem is fundamentally about demonstrating that you can make substitutions and perform the corresponding arithmetic correctly. When making a substitution, it is helpful to be in the habit of putting that substitution inside of parentheses. Sometimes it's necessary, and sometimes it's extraneous. Over time, you will have the experience and insight to see whether it's necessary, but to start off it's best be in the habit of using parentheses all the time.

Here is a presentation for finding the value of \(3a - 2b\) when \(a = 2\) and \(b = 4\text{:}\)

\begin{equation*} \begin{aligned} 3a - 2b \amp = 3(2) - 2(4) \amp \eqnspacer \amp \text{Substitute $a = 2$ and $b = 4$} \\ \amp = 6 - 8 \amp \amp \text{Arithmetic} \\ \amp = -2 \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}
\begin{equation*} \text{If and then } \end{equation*}

Try it!

Determine the value of the expression \(2x + 5y\) when \(x = 3\) and \(y = -2\text{.}\) Show the calculation and write your result as an if-then statement.

Solution.
\begin{equation*} \begin{aligned} 2x + 5y \amp = 2(3) + 5(-2) \amp \amp \text{Substitute $x = 3$ and $y = -2$} \\ \amp = 6 + (-10) \amp \amp \text{Arithmetic} \\ \amp = -4 \amp \amp \text{Arithmetic} \end{aligned} \end{equation*}
\begin{equation*} \text{If $x = 3$ and $y = -2$, then $2x + 5y = -4$.} \end{equation*}

A major step in your mathematical thinking is the ability to apply previous results in new settings, and combining old ideas in new ways. We have talked about substituting a value for a variable, and we've talked about solving for a variable. We can combine the two into a new type of problem.

Activity 5.1.2. Substituting a Value for a Variable and Solving.

Here is an example of combining the two ideas together. Consider the equation \(3m + 5n = 30\text{.}\) This equation has two variables in it. If we declare a value for one of the variables, then we can solve for the other one. For example, if we are told that \(m = 5\text{,}\) then we have the following:

\begin{equation*} \begin{aligned} 3m + 5n \amp = 30 \\ 3(5) + 5n \amp = 30 \amp \eqnspacer \amp \text{Substitute $m = 5$} \\ 15 + 5n \amp = 30 \amp \amp \text{Arithmetic} \\ 5n \amp = 15 \amp \amp \text{Subtract $15$ from both sides} \\ n \amp = 3 \amp \amp \text{Divide both sides by 5} \end{aligned} \end{equation*}

Try it!

Solve the equation \(3m + 5n = 30\) for when \(n = -2\text{.}\) Use a complete presentation.

Solution.
\begin{equation*} \begin{aligned} 3m + 5n \amp = 30 \\ 3m + 5(-2) \amp = 30 \amp \amp \text{Substitute $n = -2$} \\ 3m + (-10) \amp = 30 \amp \amp \text{Arithmetic} \\ 3m \amp = 40 \amp \amp \text{Add $10$ to both sides} \\ m \amp = \frac{40}{3} \amp \amp \text{Divide both sides by $3$} \end{aligned} \end{equation*}

There are other types of substitutions that we can make. In the definition of a variable (Definition 2.1.1), we said that variables can also represent mathematical expressions. Fortunately, there is no conceptual difference between the two. It all comes down to the proper execution of algebra.

Activity 5.1.3. Substituting an Expression for a Variable.

When substituting an expression for a variable, it is important to wrap the expression inside of parentheses. This small detail is extremely important because it represents a container that holds the entire contents of the variable. Failure to do so will often lead to failing to properly apply the distributive property or committing other types of errors.

This is how it looks to substitute \(x = 2y - 1\) into the expression \(-x + 6\text{:}\)

\begin{equation*} \begin{aligned} -x + 6 \amp = -(2y - 1) + 6 \amp \eqnspacer \amp \text{Substitute $x = 2y - 1$} \\ \amp = -2y + 1 + 6 \amp \amp \text{Distributive property} \\ \amp = -2y + 7 \amp \amp \text{Combine like terms} \end{aligned} \end{equation*}

Try it!

Substitute \(a = -2b + 3\) into the expression \(-2a - 5\) and simplify the result. Use a presentation.

Solution.
\begin{equation*} \begin{aligned} -2a - 5 \amp = -2(-2b + 3) - 5 \amp \amp \text{Substitute $a = -2b + 3$} \\ \amp = 4b - 6 - 5 \amp \amp \text{Distributive property} \\ \amp = 4b - 11 \amp \amp \text{Combine like terms} \end{aligned} \end{equation*}

Activity 5.1.4. Substituting an Expression for a Variable and Solving.

Conceptually, substituting into an expression is not different from substituting into an equation because an equation is just two expressions that are claimed to represent the same value.

Try it!

Substitute \(x = 2y + 1\) into the equation \(3x - 2y = 7\) and solve for Use a complete presentation.

Solution.
\begin{equation*} \begin{aligned} 3x - 2y \amp = 7 \\ 3(2y + 1) - 2y \amp = 7 \amp \amp \text{Substitute $x = 2y + 1$} \\ 6y + 3 - 2y \amp = 7 \amp \amp \text{Distributive property} \\ 4y + 3 \amp = 7 \amp \amp \text{Combine like terms} \\ 4y \amp = 4 \amp \amp \text{Subtract $6$ from both sides} \\ y \amp = 1 \amp \amp \text{Divide both sides by $4$} \end{aligned} \end{equation*}