Translating and Solving Applications

Most of the time a question that requires an algebraic solution comes out of a real life question. To begin with that question is asked in English (or the language of the person asking) and not in math symbols. Because of this, it is an important skill to be able to translate an everyday situation into algebraic language.

We will start by restating the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve. When assigning a variable, choose a letter that reminds you of what you are looking for. For example, you might use q for the number of quarters if you were solving a problem about coins.

Example: How to Solve Translate and Solve Applications

The MacIntyre family recycled newspapers for two months. The two months of newspapers weighed a total of 57 pounds. The second month, the newspapers weighed 28 pounds. How much did the newspapers weigh the first month?

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains text and algebra. In the top row, the first cell says “Step 1. Read the problem. Make sure all the words and ideas are understood.” The text in the second cell says “The problem is about the weight of newspapers.” The third cell is blank.

In the second row, the first cell says “Step 2. Identify what we are asked to find.” The second cell says “What are we asked to find?” The third cell says: “How much did the newspapers weigh the 2nd month?”

In the third row, the first cell says “Step 3. Name what we are looking for. Choose a variable to represent that quantity.” The second cell says “Choose a variable.” The third cell says “Let w equal weight of the newspapers the 1st month.”

In the fourth row, the first cell says “Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.” The second cell says “Restate the problem. We know that the weight of the newspapers the second month is 28 pounds.” The third cell says “Weight of newspapers the 1st month plus the weight of the newspapers the 2nd month equals 57 pounds. Weight from 1st month plus 28 equals 57.” One line down, the second cell says “Translate into an equation using the variable w.” The third cell contains the equation w plus 28 equals 57.

In the fifth row, the first cell says “Step 5. Solve the equation using good algebra techniques.” The second cell says “Solve.” The third cell contains the equation with 28 being subtracted from both sides: w plus 28 minus 28 equals 57 minus 28, with minus 28 written in red. Below this is w equals 29.

In the sixth row, the first cell says “Step 6. Check the answer and make sure it makes sense.” The second cell says “Does 1st month’s weight plus 2nd month’s weight equal 57 pounds?” The third cell contains the equation 29 plus 28 might equal 57. Below this is 57 equals 57 with a check mark next to it.

In the seventh and final row, the first cell says ‘Step 7. Answer the question with a complete sentence.” The second cell says “Write a sentence to answer ‘How much did the newspapers weigh the 2nd month?’” The third cell contains the sentence “The 2nd month the newspapers weighed 29 pounds.”

Solving an application:

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Example

Randell paid $28,675 for his new car. This was $875 less than the sticker price. What was the sticker price of the car?

Solution

\(\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & & & \\ \\ \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{“What was the sticker price of the car?”}\hfill \\ \\ \\ \mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \\ \text{Choose a variable to represent that quantity.}\hfill & & & & & \text{Let}\phantom{\rule{0.2em}{0ex}}s=\text{the sticker price of the car}.\hfill \\ \\ \\ \mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation. Restate}\hfill & & & & & \\ \text{the problem in one sentence.}\hfill & & & & & \text{\$28,675 is \$875 less than the sticker price}\hfill \\ \\ \\ & & & & & \text{\$28,675 is \$875 less than}\phantom{\rule{0.2em}{0ex}}s\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill & & & & & \hfill \begin{array}{ccc}\hfill 28,675& =\hfill & s-875\hfill \\ \hfill 28,675+875& =\hfill & s-875+875\hfill \\ \hfill 29,550& =\hfill & s\hfill \end{array}\hfill \\ \\ \\ \mathbf{\text{Step 6. Check}}\phantom{\rule{0.2em}{0ex}}\text{the answer.}\hfill & & & & & \\ \text{Is \$875 less than \$29,550 equal to \$28,675?}\hfill & & & & & \\ \hfill \begin{array}{ccc}\hfill 29,550-875& \stackrel{?}{=}\hfill & 28,675\hfill \\ \hfill 28,675& =\hfill & 28,675✓\hfill \end{array}\hfill & & & & & \\ \\ \\ \begin{array}{c}\mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question with}\hfill \\ \text{a complete sentence.}\hfill \end{array}\hfill & & & & & \text{The sticker price of the car was \$29,550.}\hfill \end{array}\)

Translating and Solving Applications