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Lesson 4 of 13

Finding the Unknown · Lesson 4 of 13

Solving Equations One Step at a Time

“Isolate an unknown through justified steps and confirm the result.”

Learning Objectives

• Solve equations by undoing operations in a sensible order. • Handle negative additions and fractional solutions. • Justify the shortcut of writing a term on the other side. • Check an answer in the original equation.

Work towards one copy of the unknown

To solve an equation, aim for a statement such as x = 3. If the unknown has been multiplied and then had a number added, first remove the added number, then undo the multiplication. This order exposes the unknown without changing its value. Writing one operation per line makes the reasoning easy to follow and helps you catch sign errors.

Undo subtraction, then multiplication

Problem
Solve 5x − 4 = 7.

  1. 1.Add 4 to both sides: 5x − 4 + 4 = 7 + 4.
  2. 2.Simplify to 5x = 11. Divide both sides by 5 to get x = 11/5.
  3. 3.Check the original equation: 5 × (11/5) − 4 = 11 − 4 = 7. Therefore x = 11/5.
Subtract a negative addition

Problem
Solve 11y + (−5) = 61.

  1. 1.The term added to 11y is −5. Undo it by subtracting −5 from both sides, which is the same as adding 5.
  2. 2.11y + (−5) − (−5) = 61 − (−5), so 11y = 66.
  3. 3.Divide by 11: y = 6. Check: 11 × 6 − 5 = 61.
Definition
Isolate the unknown

Use equal operations on both sides until the unknown stands alone.

Quotients and negative answers

An unknown divided by a known number is undone by multiplication. A negative answer does not signal a mistake: the equation may need it to balance. Fractions are also ordinary numbers. Decide whether an answer is suitable by checking the equation and, for a real situation, its meaning.

Undo division

Problem
Solve u/15 = 6.

  1. 1.Multiply both sides by 15: (u/15) × 15 = 6 × 15.
  2. 2.The left side simplifies to u, giving u = 90.
  3. 3.Check: 90/15 = 6. The 15 was a divisor, so subtraction would not undo it.
Let the answer be negative

Problem
Solve −8 = 5x − 3.

  1. 1.Add 3 to both sides: −8 + 3 = 5x − 3 + 3.
  2. 2.This gives −5 = 5x. Divide both sides by 5: −1 = x, or x = −1.
  3. 3.Check: 5 × (−1) − 3 = −5 − 3 = −8, equal to the original left side.
A negative coefficient

Problem
Solve −53w = −15.

  1. 1.The coefficient −53 means the number multiplying w. Divide both sides by −53.
  2. 2.w = (−15)/(−53) = 15/53, which is positive because the two negative signs cancel in division.
  3. 3.Check: −53 × (15/53) = −15. There is no requirement for w to be a whole number.

Understand the short form

You may see 3x − 10 = 35 followed by 3x = 35 + 10. This shortened writing comes from adding 10 to both sides and simplifying the left. It is sometimes described as moving a term across the equal sign and changing its sign. The term does not move by itself; the same-operation rule explains why the short form is valid.

The short form for a multiplying factor is different. From 3x = 45, division by 3 gives x = 45/3. The factor becomes a divisor because division undoes multiplication. It does not become −3. Always identify whether the quantity is added, subtracted, multiplied, or divided before choosing the inverse operation.

Explain every shortcut

Problem
Solve 3x − 10 = 35 and justify the short lines.

  1. 1.Write 3x = 35 + 10 = 45. This line abbreviates adding 10 to both sides.
  2. 2.Write x = 45/3 = 15. This line abbreviates dividing both sides by 3.
  3. 3.Check: 3 × 15 − 10 = 45 − 10 = 35. Each short line is supported by an equal operation.
Common mistake

Check the original equation, not just your final simplified line. An earlier sign mistake may produce an answer that fits the mistaken line. Also, a factor such as 5 in 5x is undone by division, not by subtracting 5.

Quiz

Quick check

For 7x + 9 = 30, which first step isolates the multiplied term?

Quick check

What is the solution of t/8 = −3?

Quick check

Which explains 2x − 6 = 12 becoming 2x = 18?

Quick check

What solves 4x + 3 = 5?

Quick check

Which check correctly tests x = −1 in −8 = 5x − 3?

Practice Problems

Practice Problems
  1. Solve and check 2y = 60.
  2. Solve and check 13 − z = 8. Explain how you deal with the minus sign before z.
  3. Solve 3x + 7 = −2 with one operation per line.
  4. Solve a/4 − 3 = 5.
  5. Solve 7 − 2b = 12. Explain why a negative answer is allowed.
  6. For 6r + 5 = 23, write both a full same-operation solution and a shortened solution. Match corresponding steps.
  7. A learner changes 5v = 20 to v = 20 − 5. Explain the operation error and correct it.

Key Takeaways

Key Takeaways

• Isolate the unknown by undoing the operations around it. • Undo an addition before dividing a multiplied term. • Subtracting a negative number adds its positive opposite. • Negative and fractional numbers can be solutions. • Short notation is valid only when supported by equal operations and an original-equation check.

Previous · Lesson 3

Trial, Error, and Preserving Equality

Next · Lesson 5

Unknowns on Both Sides