# 3.2: Two-Step Equations

**At Grade**Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

- Solve a two step equation using addition, subtraction, multiplication, and division.
- Solve a two-step equation by combining like terms.
- Solve real-world problems using two-step equations.

## Vocabulary

Terms introduced in this lesson:

- two-step equations
- like terms
- combining like terms
- unknown
- equation in two variables

## Teaching Strategies and Tips

Use Example 1 to motivate the process of solving two-step equations:

- Two marbles can be removed from
*each*pan first. - The first step in solving two-step equations is to move the constant
*away*from the variable term. - Whereas it was not possible before, the marbles can now be divided into three groups.
- The second step in solving two-step equations is to isolate the variable by
*dividing*by its coefficient. - It is a small jump for students to write an algebraic expression based on the equality implied by the pans and solve it in an analogous way.

To keep the pans in equilibrium, Example 1 also teaches that, “what is done to one side must be done to the other side”.

Because the solution to Example 2 is negative, the balance strategy of Example 1 will not apply. Use a similar problem in which the solution is positive to demonstrate the balance strategy for *variables buried in parentheses*.

Example.

*Six bags each containing the same unknown number of blue marbles and \begin{align*}1\end{align*} 1 red marble are placed on one side of a balance. \begin{align*}12\end{align*}12 red marbles are put on the other side. The scales balance. How many blue marbles are in each bag? Assume the marbles weigh the same and the bags weigh nothing*.

Solution. The unknown quantity is *the number of blue marbles in each bag* and is denoted by \begin{align*}x\end{align*}

\begin{align*}6(x+1)=12\end{align*}

This is an example of an equation where \begin{align*}x\end{align*} is *buried in parentheses*. To find the number of blue marbles, proceed in one of two ways: (1) Observe that \begin{align*}1\end{align*} bag weighs the same as \begin{align*}2\end{align*} red marbles. This is equivalent to dividing both sides of the above equation by \begin{align*}6\end{align*}. (2) Empty the contents of each bag onto the pan. There will be \begin{align*}6\end{align*} red marbles and \begin{align*}6\;\mathrm{times}\end{align*} the number of blue marbles that was in one bag. This is equivalent to distributing the \begin{align*}6\end{align*} in the above equation. In both approaches, the equations are reduced to the familiar one-step and two-step equations, respectively.

The first approach is easier since \begin{align*}12\end{align*} is evenly divided by \begin{align*}6\end{align*}.

\begin{align*}\frac {\cancel{6}(x+1)}{\cancel{6}} & = \frac{12}{6}\\ x+\cancel{1} & = 2\\ -\cancel{1} & = -1\\ x & = 1\end{align*}

General Tip: The first step in solving equations with *variables buried in parentheses* depends on:

- Whether the constant is evenly divisible by the coefficient. See Example 2.
- Whether fractions are present. See Examples 3 and 4.

Warm-up to Examples 5 and 6 with exercises similar to the following:

*Which of the following pairs of expressions are like terms?*

- \begin{align*}x\end{align*} and \begin{align*}5x\end{align*} (Yes.)
- \begin{align*}x\end{align*} and \begin{align*}xy\end{align*} (No.)
- \begin{align*}x\end{align*} and \begin{align*}x^2\end{align*} (No.)
- \begin{align*}-11\end{align*} and \begin{align*}11\end{align*} (Yes.)

In Examples 8 and 9, two-variable equations will result. Students substitute one of the givens for one variable to determine the other.

## Error Troubleshooting

Watch for the switch from Example 9(ii) to 9(iii), *from Celsius to Fahrenheit*.

*Review Question* 1c. Remind students to distribute the negative to *both* terms in the parentheses.

*Review Question* 1g. Hint: Write the \begin{align*}s\end{align*} term with a common denominator as \begin{align*}\frac{8}{8}s\end{align*}.