Everything covered in this lesson, in one place - useful for revision or printing.
\( 3x + 7 = 12 \)
Lesson 1 covered balance, inverse operations and two-step equations. This lesson works through every method you need for the harder equations on your worksheet.
You will learn how to handle:
\( \text{do the same to BOTH sides} \)
An equation is a balance. Whatever you do to the left, you must do to the right, or it is no longer the equation you were given.
\( 3x + 7 = 12 \)
The \(x\) is trapped by two operations: multiplied by 3, then 7 added.
Step 1 — undo the \(+7\): subtract 7 from both sides, giving \(3x = 5\).
Step 2 — undo the \(\times 3\): divide both sides by 3.
\( 4x - 5 - x = 10 \)
If one side has several \(x\) terms, tidy that side before anything else.
\(4x - x = 3x\), so the equation becomes \(3x - 5 = 10\).
\( 2(x + 3) = 18 \)
Multiply everything inside the bracket by the number outside — both terms, not just the first.
\(2x + 6 = 18\), so \(2x = 12\) and \(x = 6\).
\( 5x - 3(2x + 1) = 6 \)
The bracket is multiplied by negative 3, so every term inside changes sign.
\(-3 \times 2x = -6x\) and \(-3 \times +1 = -3\).
\( 7(x - 2) = 3(2x + 1) \)
Expand each side separately first: \(7x - 14 = 6x + 3\).
Now move the smaller \(x\) term across, so the \(x\) you keep stays positive.
\( 4(x + 5) = 2(x - 3) \)
Expand: \(4x + 20 = 2x - 6\). Watch that \(2 \times -3 = -6\).
Move \(2x\) left and \(20\) right — each change of side flips the sign.
\( 7(x - 9) = 10x \)
Expanding gives \(7x - 63 = 10x\). Here \(10x\) is the bigger term, so collect on the right.
\(-63 = 3x\), so \(x = -21\).
\( \dfrac{x}{3} + 4 = 10 \)
Deal with the whole number first, then multiply both sides by the denominator.
Subtract 4: \(\dfrac{x}{3} = 6\). Multiply both sides by 3: \(x = 18\).
\( \dfrac{p + 3}{5} - 8 = 1 \)
The entire \(p + 3\) sits on top of the 5 as one package, so it all gets multiplied by 5.
Add 8: \(\dfrac{p+3}{5} = 9\). Multiply by 5: \(p + 3 = 45\).
\( \dfrac{y}{3} + \dfrac{y + 1}{2} = 7 \)
With more than one denominator, find the lowest common denominator and multiply every term by it.
Here the LCD is 6, including the \(7\) on the right.
\( \dfrac{1}{2}(q + 3) - 4 = \dfrac{1}{4}(q - 3) \)
Clear the fractions first — the brackets are far easier without them. Denominators 2 and 4 give an LCD of 4.
\( \dfrac{4}{x} + 2 = 8 \)
Isolate the fraction first, then multiply by \(x\). Do not try to expand anything.
Subtract 2: \(\dfrac{4}{x} = 6\). Multiply both sides by \(x\): \(4 = 6x\).
\( \dfrac{1}{x} + 8 = 3 \)
Same method. Isolating the fraction leaves a negative on the right.
\(\dfrac{1}{x} = -5\), so \(1 = -5x\) and \(x = -\dfrac{1}{5}\).
\( m(m + 3) - 7 = m^2 + 2m - 1 \)
Do not panic at the \(m^2\). Expand both sides and see what survives.
\(m^2 + 3m - 7 = m^2 + 2m - 1\). The \(m^2\) is on both sides, so it subtracts away entirely.
\( (m + 3)(m - 6) = (m - 4)(m - 7) \)
Expand each pair of brackets in full before comparing.
\(m^2 - 3m - 18 = m^2 - 11m + 28\). Take the double negatives slowly: \(-4 \times -7 = +28\).
\( t(t + 4) - 5t = t(6 + t) - 8 \)
Expand everything: \(t^2 + 4t - 5t = 6t + t^2 - 8\).
Tidy the left: \(t^2 - t = t^2 + 6t - 8\). The \(t^2\) cancels.
\( 3\left(\dfrac{5}{3}\right) + 7 = 12 \)
Substitute your answer into the original equation — not into a line you rewrote, in case that line already has the mistake.
\( \text{brackets} \to \text{fractions} \to \text{like terms} \to \text{move } x \to \text{divide} \)
Not every equation needs every step, but this order never fails:
\( 3x = 5 \;\Rightarrow\; x = \dfrac{5}{3} \)
\( \text{expand} \to \text{clear} \to \text{collect} \to \text{divide} \)