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Lesson 7 of 13

Finding the Unknown · Lesson 7 of 13

Equations for Everyday Situations

“Choose an unknown carefully and model costs, savings, and grouped operations.”

Learning Objectives

• Choose and define an unknown for a word problem. • Write equivalent equations using different unknown quantities. • Find when two savings totals become equal. • Compare three valid methods for a grouped equation.

Choose what the letter counts

Before writing an equation, decide exactly what your letter represents. Suppose a meal costs ₹25 per person with a single ₹50 delivery charge. There is a ₹500 budget and five family members will attend. A letter for all people and a letter for invited friends count different things, so they need different expressions. The delivery fee is added once, not once per person.

Count everybody first

Problem
With the whole ₹500 budget, how many friends can join five family members?

  1. 1.Let p be the total number of people. Total cost is 25p + 50, so 25p + 50 = 500.
  2. 2.Subtract 50: 25p = 450. Divide by 25: p = 18 people.
  3. 3.The 18 include five family members, so invited friends = 18 − 5 = 13.
  4. 4.Check: 25 × (13 + 5) + 50 = 450 + 50 = ₹500. A nineteenth person would raise the cost above the budget.
Choose friends as the unknown

Problem
Solve the same party problem with f representing only the friends.

  1. 1.The total number of people is f + 5. Thus 25(f + 5) + 50 = 500.
  2. 2.Subtract 50: 25(f + 5) = 450. Divide by 25: f + 5 = 18.
  3. 3.Subtract 5: f = 13. This directly gives the requested number of friends.
  4. 4.Both equations describe the same party. Their unknowns differ, with p = f + 5.

You can also solve by arithmetic: reserve ₹50 for delivery, divide the remaining ₹450 by ₹25, then remove the five family places. The equation records these relationships in a form that helps when the unknown appears in more than one part of a problem.

Two quantities become equal

Savings problems often have an initial amount and a repeated addition. If m counts completed months, a person starting with ₹4000 and adding ₹650 each month has 4000 + 650m. This expression describes a total, not just the new savings. Set two such expressions equal when the question asks when the totals match.

Meet at the same savings total

Problem
One saver starts with ₹4000 and adds ₹650 per month. Another starts with ₹5050 and adds ₹500 per month. When are their savings equal?

  1. 1.Let m be the number of months. Write 4000 + 650m = 5050 + 500m.
  2. 2.Subtract 500m: 4000 + 150m = 5050. Subtract 4000: 150m = 1050.
  3. 3.Divide by 150: m = 7 months.
  4. 4.Check: 4000 + 650 × 7 = ₹8550 and 5050 + 500 × 7 = ₹8550. The initial gap ₹1050 closes by ₹150 each month.
Initial gap: ₹1050First saver adds ₹150 more each month₹1050 ÷ ₹150 per month = 7 monthsAt month 7: ₹8550 each
The savings gap closes steadily— The difference in monthly additions, rather than either addition alone, closes the initial gap.

One equation, three sensible methods

An equation with brackets can be solved without expanding if undoing the outer operations is easy. Expanding is another valid route. You can also divide the entire equation by a convenient common factor. Comparing the routes shows that an equation does not demand one fixed sequence; each route must preserve equality and make the work simpler.

Undo the outer operations

Problem
Solve 28(x + 4) + 300 = 1000 by working from outside the brackets.

  1. 1.Subtract 300 from both sides: 28(x + 4) = 700.
  2. 2.Divide by 28: x + 4 = 25. Subtract 4: x = 21.
  3. 3.Check: 28(21 + 4) + 300 = 28 × 25 + 300 = 700 + 300 = 1000.
Expand first

Problem
Solve 28(x + 4) + 300 = 1000 by distributing 28.

  1. 1.Expand every bracket term: 28x + 112 + 300 = 1000.
  2. 2.Combine constants: 28x + 412 = 1000. Subtract 412: 28x = 588.
  3. 3.Divide by 28: x = 21. Substituting in the original again gives 1000.
Reduce the numbers first

Problem
Solve 28(x + 4) + 300 = 1000 by dividing the entire equation by 4.

  1. 1.Divide both sides, including every left-side term, by 4: 7(x + 4) + 75 = 250.
  2. 2.Subtract 75: 7(x + 4) = 175. Expand: 7x + 28 = 175.
  3. 3.Subtract 28: 7x = 147. Divide by 7: x = 21. Every quantity divided by 4 had an integer result, making the numbers smaller without altering the solution.
Common mistake

A letter for friends excludes family members; a letter for total people includes them. When dividing an equation such as 28(x + 4) + 300 = 1000, divide the +300 term too. Leaving it unchanged creates a different equation.

Quiz

Quick check

If f counts friends and five family members attend, what counts all people?

Quick check

Which party cost equation uses a once-only ₹50 delivery fee?

Quick check

How much faster does the first saver’s total grow when monthly additions are ₹650 and ₹500?

Quick check

After subtracting 300 from 28(x + 4) + 300 = 1000, what remains?

Quick check

What results from dividing every term of 28(x + 4) + 300 = 1000 by 4?

Practice Problems

Practice Problems
  1. A snack costs ₹30 per person and delivery is ₹60. A ₹600 budget includes six family members. Find the maximum number of friends if the whole budget can be used.
  2. Write both a total-people equation and a friends-only equation for that snack problem.
  3. One saver has ₹1200 and adds ₹200 monthly; another has ₹1800 and adds ₹100 monthly. Find when their totals match and calculate the common total.
  4. Solve 12(x + 3) + 24 = 120 by undoing outer operations and by expanding. Compare the work.
  5. Solve 18(y + 2) + 90 = 450 by first dividing the whole equation by 9.
  6. Explain why 25f + 5 + 50 = 500 does not correctly represent the party when f counts friends.

Key Takeaways

Key Takeaways

• Define the unknown before translating the story. • A repeated cost is multiplied; a once-only fee is added once. • Different unknown choices can produce equivalent models. • Compare complete savings totals, including starting amounts. • Several solving methods work when every step preserves equality.

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Equations Hidden in Patterns

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Number Tricks and Equation Chains