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Solving one- and two-step equations

Algebra · level 4, lesson 4 of 7

A question to start

A balance scale holds 3 identical bags and 5 single marbles on the left, and 20 marbles on the right. It balances perfectly. How many marbles are in each bag?

You can probably work it out: take 5 marbles off each side, so 3 bags balance 15 marbles; then each bag holds 5. What you just did, written in algebra, is

\[3x + 5 = 20 \quad\Rightarrow\quad 3x = 15 \quad\Rightarrow\quad x = 5.\]

This lesson turns that balance thinking into a method that solves any one- or two-step equation, including ones with negative numbers and fractions, where a real balance scale cannot help you any more.

The idea

What solving means

An equation says two expressions are equal, such as \(3x + 5 = 20\). A solution is a value of the letter that makes the equation true. \(x = 5\) is a solution because \(3 \times 5 + 5 = 20\) is true. \(x = 4\) is not, because \(3 \times 4 + 5 = 17\), not 20. Solving means finding the solution.

The balance rule

An equation is like a balanced scale: the left side "weighs" the same as the right side.

xxx 3x + 5 20

Whatever you do to one side, do exactly the same to the other side, and the equation stays true.

Why is this rule safe? If two numbers are equal, they are the same number written in two ways. Adding 7 to the same number gives the same answer, whichever way it was written. The same goes for subtracting, multiplying, or dividing (by anything except zero). So equal things stay equal.

Inverse operations: the undo buttons

To get the letter on its own, you remove what has been done to it, using the inverse (opposite) operation:

Done to \(x\) Undo it by
\(+ 7\) \(- 7\) from both sides
\(- 4\) \(+ 4\) to both sides
\(\times 3\) \(\div 3\) on both sides
\(\div 5\) \(\times 5\) on both sides

One-step equations

\(x + 7 = 12\): subtract 7 from both sides. \(x = 5\).

\(\dfrac{x}{4} = 6\): multiply both sides by 4. \(x = 24\).

Write each step on a new line, with the equals signs lined up, and say what you did to both sides:

\[ \begin{aligned} x + 7 &= 12\\ x + 7 - 7 &= 12 - 7 \qquad \text{(subtract 7 from both sides)}\\ x &= 5 \end{aligned} \]

Two-step equations: undo in reverse order

In \(3x - 5 = 16\), two things happened to \(x\): first it was multiplied by 3, then 5 was subtracted. Like taking off shoes before socks, undo the last thing first:

\[ \begin{aligned} 3x - 5 &= 16\\ 3x &= 21 \qquad \text{(add 5 to both sides)}\\ x &= 7 \qquad \text{(divide both sides by 3)} \end{aligned} \]

This is the two-step machine from level 3, written as an equation. The order matters for the same reason: the "\(-5\)" is on the outside, so it comes off first.

Always check

Substitute your answer into the original equation. For \(x = 7\): \(3 \times 7 - 5 = 21 - 5 = 16\). ✓ If the check fails, there is a mistake somewhere, and you know to look for it.

Equations that need negatives and fractions

The balance rule works with any numbers, not only the ones a real scale can hold:

  • \(x + 9 = 4\): subtract 9 from both sides, \(x = -5\). Check: \(-5 + 9 = 4\). ✓
  • \(2x = 7\): divide both sides by 2, \(x = \dfrac{7}{2} = 3.5\). Check: \(2 \times 3.5 = 7\). ✓
  • \(-4x = 20\): divide both sides by \(-4\), \(x = -5\). Check: \(-4 \times (-5) = 20\). ✓

When the letter is subtracted

In \(10 - x = 3\), the \(x\) is being taken away. A neat way is to add \(x\) to both sides first, so the letter becomes positive: \(10 = 3 + x\). Then subtract 3: \(x = 7\). Check: \(10 - 7 = 3\). ✓

Worked examples

1. One step. Solve \(x - 8 = 13\).

Add 8 to both sides: \(x = 21\). Check: \(21 - 8 = 13\). ✓

2. Two steps. Solve \(4m + 9 = 37\).

  • Subtract 9 from both sides: \(4m = 28\).
  • Divide both sides by 4: \(m = 7\).
  • Check: \(4 \times 7 + 9 = 28 + 9 = 37\). ✓

3. A fraction. Solve \(\dfrac{x}{3} + 4 = 10\).

  • Subtract 4 from both sides: \(\dfrac{x}{3} = 6\).
  • Multiply both sides by 3: \(x = 18\).
  • Check: \(18 \div 3 + 4 = 6 + 4 = 10\). ✓

4. A negative answer. Solve \(5y + 12 = 2\).

  • Subtract 12 from both sides: \(5y = -10\).
  • Divide both sides by 5: \(y = -2\).
  • Check: \(5 \times (-2) + 12 = -10 + 12 = 2\). ✓

5. Simplify first. Solve \(2x + 3x - 4 = 21\).

  • Collect like terms: \(5x - 4 = 21\).
  • Add 4: \(5x = 25\). Divide by 5: \(x = 5\).
  • Check: \(10 + 15 - 4 = 21\). ✓

6. A letter being subtracted. Solve \(20 - 3n = 8\).

  • Add \(3n\) to both sides: \(20 = 8 + 3n\).
  • Subtract 8: \(12 = 3n\). Divide by 3: \(n = 4\).
  • Check: \(20 - 12 = 8\). ✓

Try it

  1. Solve \(x + 15 = 41\).
  2. Solve \(\dfrac{k}{6} = 7\).
  3. Solve \(2x + 11 = 35\).
  4. Solve \(7p - 3 = 46\).
  5. Solve \(\dfrac{a}{5} - 2 = 4\).
  6. Solve \(3w + 20 = 5\).
  7. Solve \(15 - 2y = 1\).
  8. Sam solved \(4x + 6 = 30\) like this: "\(4x = 30 - 6 = 24\), so \(x = 24 - 4 = 20\)." Find his mistake and solve it correctly.
Answers
  1. Subtract 15: \(x = 26\). Check: \(26 + 15 = 41\). ✓
  2. Multiply by 6: \(k = 42\). Check: \(42 \div 6 = 7\). ✓
  3. Subtract 11: \(2x = 24\). Divide by 2: \(x = 12\). Check: \(24 + 11 = 35\). ✓
  4. Add 3: \(7p = 49\). Divide by 7: \(p = 7\). Check: \(49 - 3 = 46\). ✓
  5. Add 2: \(\dfrac{a}{5} = 6\). Multiply by 5: \(a = 30\). Check: \(30 \div 5 - 2 = 4\). ✓
  6. Subtract 20: \(3w = -15\). Divide by 3: \(w = -5\). Check: \(3 \times (-5) + 20 = -15 + 20 = 5\). ✓
  7. Add \(2y\): \(15 = 1 + 2y\). Subtract 1: \(14 = 2y\), so \(y = 7\). Check: \(15 - 14 = 1\). ✓
  8. \(4x\) means 4 times \(x\), so the undo is divide by 4, not subtract 4. \(4x = 24\) gives \(x = 6\). Check: \(4 \times 6 + 6 = 30\). ✓ (His answer 20 gives \(86\), not 30.)

Common mistake

Doing something to one side only. Crossing out the \(+5\) in \(3x + 5 = 20\) without taking 5 from the 20 tips the scale. Every step must happen on both sides.

Using the wrong inverse. \(4x\) is undone by dividing by 4; \(x + 4\) is undone by subtracting 4. Read the expression aloud ("four times \(x\)") before choosing the undo button.

Where it leads