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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.”
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Translate into an algebraic equation and solve:

The Pappas family has two cats, Zeus and Athena. Together, they weigh 23 pounds. Zeus weighs 16 pounds. How much does Athena weigh?

7 pounds

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Translate into an algebraic equation and solve:

Sam and Henry are roommates. Together, they have 68 books. Sam has 26 books. How many books does Henry have?

42 books

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Solve an application.

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

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

Step 1. Read the problem. Step 2. Identify what we are looking for. “What was the sticker price of the car?” Step 3. Name what we are looking for. Choose a variable to represent that quantity. Let s = the sticker price of the car . Step 4. Translate into an equation. Restate the problem in one sentence. $28,675 is $875 less than the sticker price $28,675 is $875 less than s Step 5. Solve the equation. 28,675 = s − 875 28,675 + 875 = s − 875 + 875 29,550 = s Step 6. Check the answer. Is $875 less than $29,550 equal to $28,675? 29,550 − 875 = ? 28,675 28,675 = 28,675 ✓ Step 7. Answer the question with a complete sentence. The sticker price of the car was $29,550.

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Translate into an algebraic equation and solve:

Eddie paid $19,875 for his new car. This was $1,025 less than the sticker price. What was the sticker price of the car?

$20,900

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Translate into an algebraic equation and solve:

The admission price for the movies during the day is $7.75. This is $3.25 less the price at night. How much does the movie cost at night?

$11.00

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Key concepts

  • To Determine Whether a Number is a Solution to an Equation
    1. Substitute the number in for the variable in the equation.
    2. Simplify the expressions on both sides of the equation.
    3. Determine whether the resulting statement is true.
      • If it is true, the number is a solution.
      • If it is not true, the number is not a solution.
  • Addition Property of Equality
    • For any numbers a , b , and c , if a = b , then a + c = b + c .
  • Subtraction Property of Equality
    • For any numbers a , b , and c , if a = b , then a − c = b − c .
  • To Translate a Sentence to an Equation
    1. Locate the “equals” word(s). Translate to an equal sign (=).
    2. Translate the words to the left of the “equals” word(s) into an algebraic expression.
    3. Translate the words to the right of the “equals” word(s) into an algebraic expression.
  • To Solve an Application
    1. Read the problem. Make sure all the words and ideas are understood.
    2. Identify what we are looking for.
    3. Name what we are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation. It may be helpful to restate the problem in one sentence with the important information.
    5. Solve the equation using good algebra techniques.
    6. Check the answer in the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

Questions & Answers

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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