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3 Algebra

Syllabus
2024
Topic
3
Level
—

Choose the symbol that states the relationship

An algebraic symbol states a precise relationship between quantities. Read the complete statement from left to right and keep each symbol's meaning separate from equality.

Symbol Meaning Example and reading
<< less than 8 m<12 m8\,\mathrm{m}<12\,\mathrm{m}: 8 m is less than 12 m
>> greater than 5 s>3 s5\,\mathrm{s}>3\,\mathrm{s}: 5 s is greater than 3 s
∝\propto proportional to F∝aF\propto a: for fixed mass, F/aF/a is constant
∼\sim approximately 9.8 N/kg∼10 N/kg9.8\,\mathrm{N/kg}\sim10\,\mathrm{N/kg}

A proportionality becomes an equation only after a constant is introduced: y∝xy\propto x means y=kxy=kx for constant kk. If xx doubles under the same conditions, yy doubles. An inverse relationship must be written explicitly, such as y∝1/xy\propto1/x.

Do not read ∝\propto as “equals”; the constant and its unit may be essential. The symbol ∼\sim reports an approximation, not exact equality. For inequalities, reversing the order reverses the symbol: 3<73<7 and 7>37>3 describe the same comparison.

Change the subject while keeping the equation balanced

The subject is the single symbol isolated on one side of an equation. Changing the subject rewrites the same relationship in an equivalent form; every operation used to isolate the target must preserve equality.

Choose the required subject, then undo operations in reverse order. Apply the same operation to both sides, treat a numerator or bracket as one complete expression, and simplify only after the target is isolated. A useful check is to rearrange the result back to the original form.

v2=u2+2as⇒v2−u2=2as⇒s=v2−u22av^2=u^2+2as \quad\Rightarrow\quad v^2-u^2=2as \quad\Rightarrow\quad s=\frac{v^2-u^2}{2a}

In the example, subtracting u2u^2 from both sides removes the added term; dividing both sides by 2a2a then leaves ss. The entire difference v2−u2v^2-u^2 remains in the numerator. No numerical values are needed because the job is to produce a reusable symbolic formula.

“Move it across and change the sign” hides the balancing operation and often loses brackets or factors. Addition is undone by subtraction, multiplication by division, squaring by a square root when appropriate. Never divide by an expression that could be zero without recognising that condition.

Substitute values only after the units agree

Substitution replaces each symbol in a physical equation with its measured value. The numerical calculation is valid only when the units are compatible with the equation, so convert first and keep the value and unit together throughout the reasoning.

Write the equation → identify every symbol and required unit → convert prefixes or time units → insert values with brackets where signs or powers could be ambiguous → calculate → state the derived unit and appropriate final precision. Showing the substitution makes both the arithmetic and unit choices checkable.

E=Pt,P=60 W,t=2.0 min=120 sE=Pt,\quad P=60\,\mathrm{W},\quad t=2.0\,\mathrm{min}=120\,\mathrm{s}

Because 1 W=1 J/s1\,\mathrm{W}=1\,\mathrm{J/s}, use seconds: E=(60 J/s)(120 s)=7200 J=7.2×103 JE=(60\,\mathrm{J/s})(120\,\mathrm{s})=7200\,\mathrm{J}=7.2\times10^3\,\mathrm{J}. The seconds cancel, leaving joules. Substituting 2.0 directly would incorrectly treat minutes as seconds and make the result sixty times too small.

Units are not decorations added after a unitless calculation. Do not mix centimetres with metres, minutes with seconds, or prefixed and base units unless the equation is written for that combination. Retain full calculator precision during the calculation and round the final result rather than each intermediate value.

Solve an equation by undoing operations in order

Solving an equation finds the value of an unknown that makes both sides equal. Preserve the balance by performing the same inverse operation on both sides until the unknown is isolated.

Simplify each side if needed, undo addition or subtraction, then undo multiplication or division. Keep signs and brackets visible. When the unknown represents a physical quantity, attach the unit after the algebra and check that the value is sensible for the given quantities.

12=2+5t⇒10=5t⇒t=212=2+5t \quad\Rightarrow\quad 10=5t \quad\Rightarrow\quad t=2

The first step subtracts 2 from both sides; the second divides both sides by 5. If the equation describes time in seconds, the solution is t=2 st=2\,\mathrm{s}. Verify it by substitution: 2+5(2)=122+5(2)=12, so the found value satisfies the original equation.

Changing the subject produces a symbolic formula; solving uses the given numbers to find a particular unknown value. Do not change a sign merely because a term appears on the other side—state the inverse operation. A solution that fails when substituted back reveals an arithmetic or algebraic error.