Everything covered in this lesson, in one place - useful for revision or printing.
\( 5x + 8 = 28 \)
An equation is a number sentence with an equals sign. Solving it means finding the value that makes both sides match.
In this lesson you will learn to:
\( 6x + 4 \qquad\text{vs}\qquad 6x + 4 = 22 \)
The difference is one symbol.
\( \underbrace{6x+4}_{\text{LHS}} \;=\; \underbrace{22}_{\text{RHS}} \)
Everything before the equals sign is the left-hand side (LHS). Everything after it is the right-hand side (RHS).
\( 6x + 4 = 22 \;\Rightarrow\; x = 3 \)
The value that makes the equation true is called the solution, or the root.
\( \text{equation} \Rightarrow \text{has } = \Rightarrow \text{can be solved} \)
Key idea: no equals sign means no solution to find. If a question says "simplify" you have an expression; if it says "solve" you have an equation.
\( \text{LHS} \;=\; \text{RHS} \)
Think of an equation as a balance scale that is currently level.
\( +6 \longleftrightarrow -6 \qquad \times 5 \longleftrightarrow \div 5 \)
Every operation has an inverse that undoes it.
\( \text{do the same to BOTH sides} \)
Key idea: this single rule is every method in this lesson. Everything else is just deciding which inverse to use, and in what order.
\( x + 7 = 15 \)
For simple equations, ask yourself: what number makes this true?
\( 4m = 28 \qquad 21 - a = 9 \)
\( 5x + 8 = 28 \)
Inspection works well for one operation. With two, guessing gets unreliable.
\( \text{one step} \Rightarrow \text{inspect} \)
Key idea: use inspection for one-step equations and always justify it. The moment there are two operations, switch to inverse operations.
\( x - 6 = 2 \)
\(6\) has been subtracted from \(x\). Undo it by adding 6 — to both sides.
\( \dfrac{x}{4} = 3 \)
\(x\) has been divided by 4. Undo it by multiplying by 4 — both sides.
\( \text{one side only} \;\Rightarrow\; \text{broken} \)
\( \text{apply the inverse to BOTH sides} \)
Key idea: identify what has been done to the variable, then undo it with its inverse — on both sides, every time.
\( 5x + 8 = 28 \)
Two operations have been applied to \(x\): multiply by 5, then add 8.
\( 5x + 8 - 8 = 28 - 8 \)
Step 1 — undo the \(+8\):
\( 5(4) + 8 = 20 + 8 = 28 \;\checkmark \)
Substitute the answer back into the original equation.
\( 24 - 3b = 9 \)
\( \text{undo} + \text{and} - \text{ first, then} \times \text{and} \div \)
Key idea: subtract or add first to isolate the variable term, then divide by the coefficient. The sign travels with the term.
\( 3p - 7 = p + 9 \)
There are \(p\) terms on both sides. Collect the variables on one side and the numbers on the other.
\( 3(8) - 7 = 17 \qquad 8 + 9 = 17 \)
Work out each side separately using \(p = 8\).
\( 3(x + 4) = 21 \)
Multiply everything inside the bracket by 3:
\( \dfrac{3(x+4)}{3} = \dfrac{21}{3} \)
Because 21 divides exactly by 3, you can divide first instead:
\( \text{expand} \to \text{collect} \to \text{solve} \)
Key idea: clear brackets, gather variables on one side and numbers on the other, then finish with the two-step method.
\( x^2 = 49 \qquad x^3 = 27 \)
Ask which number multiplied by itself gives the answer.
\( 2^x = 8 \;\Rightarrow\; 2^x = 2^3 \)
When the unknown is the exponent, write both sides as powers of the same base.
\( \text{Twice a number, increased by 8, is 20.} \)
Let \(n\) be the number. Write what the sentence says, in the order it says it:
\( x + (3x - 5) = 27 \)
Two lengths add to 27 cm. The second is 5 cm shorter than three times the first.
\( \text{name it} \to \text{write it} \to \text{solve it} \to \text{check it} \)
Key idea: for powers, match the base or find the root. For word problems, name the unknown before you write anything else.
\[ \text{brackets} \to \text{collect} \to +\,- \to \times\,\div \to \text{check} \]
You now know how to: