Solve for x (ax + b = c)

\( ax+b=c\;\Rightarrow\;x=\tfrac{c-b}{a}\;(a\ne0) \)
Algebra GCSE

\( 30x + 45 = 15 \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
Subtract 45, then divide by 30.

Explanation

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Statement

In many GCSE questions you are asked to solve linear equations of the form \( ax + b = c \). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) is the unknown. To solve for \(x\), you need to rearrange the equation so that \(x\) is on its own.

\[ ax + b = c \quad \Longrightarrow \quad x = \frac{c - b}{a}, \quad (a \neq 0) \]

Why it’s true

  • The equation \(ax + b = c\) states that multiplying \(x\) by \(a\) and then adding \(b\) gives \(c\).
  • To isolate \(x\), we must reverse these operations in the opposite order: subtract \(b\) first, then divide by \(a\).
  • This process relies on the balance method: whatever you do to one side of the equation must also be done to the other.

Recipe (how to use it)

  1. Start with the equation \(ax + b = c\).
  2. Subtract \(b\) from both sides: \(ax = c - b\).
  3. Divide both sides by \(a\): \(x = \frac{c - b}{a}\).

Spotting it

You will recognise this type of equation when you see a single variable multiplied by a number, then increased or decreased by another number, equal to a result. It is the most common starting point for solving equations at GCSE foundation level.

Common pairings

  • Equations where brackets must be expanded before isolating \(x\).
  • Word problems that lead to a linear equation (e.g. "Twice a number plus 5 is 17").

Mini examples

  1. Given: \(3x + 2 = 11\). Find: \(x\). Answer: \(x = 3\).
  2. Given: \(5x - 7 = 18\). Find: \(x\). Answer: \(x = 5\).

Pitfalls

  • Forgetting to subtract \(b\) before dividing — students often divide too soon.
  • Dropping the negative sign when \(b\) or \(c\) is negative.
  • Dividing by zero — not valid, so \(a \neq 0\).
  • Arithmetic slips (e.g. \(11 - 2 = 8\) instead of \(9\)).

Exam strategy

  • Show clear steps: subtract first, then divide.
  • Check the solution by substituting back into the original equation.
  • Write fractions in simplest form unless asked for a decimal.

Summary

Linear equations of the form \(ax + b = c\) are solved by subtracting \(b\) and dividing by \(a\). This simple rearrangement appears across GCSE algebra, word problems, and practical contexts such as perimeter, area, and ratio. Mastering it ensures confidence with more complex equations later.

Worked examples

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  1. \( Solve: 3x + 5 = 20 \)
    1. \( 3x + 5 = 20 \)
    2. \( 3x = 20 - 5 \)
    3. \( 3x = 15 \)
    4. \( x = 15 / 3 \)
    Answer: \( x = 5 \)
  2. \( Solve: 7x - 4 = 24 \)
    1. \( 7x - 4 = 24 \)
    2. \( 7x = 24 + 4 \)
    3. \( 7x = 28 \)
    4. \( x = 28 / 7 \)
    Answer: \( x = 4 \)
  3. \( Solve: 2x + 9 = 3 \)
    1. \( 2x + 9 = 3 \)
    2. \( 2x = 3 - 9 \)
    3. \( 2x = -6 \)
    4. \( x = -6 / 2 \)
    Answer: \( x = -3 \)
  4. \( Solve: 9x + 11 = 65 \)
    1. \( 9x + 11 = 65 \)
    2. \( 9x = 65 - 11 \)
    3. \( 9x = 54 \)
    4. \( x = 54 / 9 \)
    Answer: \( x = 6 \)
  5. \( Solve: 4x - 15 = 9 \)
    1. \( 4x - 15 = 9 \)
    2. \( 4x = 9 + 15 \)
    3. \( 4x = 24 \)
    4. \( x = 24 / 4 \)
    Answer: \( x = 6 \)
  6. \( Solve: 6x + 2 = -10 \)
    1. \( 6x + 2 = -10 \)
    2. \( 6x = -10 - 2 \)
    3. \( 6x = -12 \)
    4. \( x = -12 / 6 \)
    Answer: \( x = -2 \)
  7. \( Solve: 15x - 8 = 67 \)
    1. \( 15x - 8 = 67 \)
    2. \( 15x = 67 + 8 \)
    3. \( 15x = 75 \)
    4. \( x = 75 / 15 \)
    Answer: \( x = 5 \)
  8. \( Solve: 12x + 19 = 7 \)
    1. \( 12x + 19 = 7 \)
    2. \( 12x = 7 - 19 \)
    3. \( 12x = -12 \)
    4. \( x = -12 / 12 \)
    Answer: \( x = -1 \)
  9. \( Solve: 25x - 30 = 95 \)
    1. \( 25x - 30 = 95 \)
    2. \( 25x = 95 + 30 \)
    3. \( 25x = 125 \)
    4. \( x = 125 / 25 \)
    Answer: \( x = 5 \)
  10. \( Solve: 18x + 45 = 9 \)
    1. \( 18x + 45 = 9 \)
    2. \( 18x = 9 - 45 \)
    3. \( 18x = -36 \)
    4. \( x = -36 / 18 \)
    Answer: \( x = -2 \)