3 Algebra
- Syllabus
- 2024
- Topic
- 3
- Level
- —
The symbols <, >, ∝ and ∼ describe different relationships and must not be used as interchangeable versions of equals.
| Symbol | Read as | Meaning | Example |
|---|---|---|---|
| a<b | a is less than b | a has the smaller value | pH<7 |
| a>b | a is greater than b | a has the larger value | temperature >25∘C |
| y∝x | y is directly proportional to x | y=kx for constant k; doubling x doubles y | mass ∝ amount for a fixed substance |
| a∼b | a is approximately or of similar scale to b | values are close, not exactly equal | 6.41∼6.4 |
For y∝x, the ratio y/x stays constant. Use < and > with the pointed end facing the smaller value.
A correlation does not establish proportionality, and ∼ does not claim exact equality. Direct proportionality must pass through the origin in a graph of y against x.
Changing the subject rewrites an equation so the required variable is alone on one side while the relationship remains equivalent.
| Operation around target | Inverse used on both sides |
|---|---|
| +a | subtract a |
| −a | add a |
| ×a | divide by a |
| ÷a | multiply by a |
| square | take the appropriate square root |
c=\frac{n}{V}\quad\Longrightarrow\quad n=cV\quad\text{and}\quad V=\frac{n}{c}
Work outwards from the target variable and undo operations in reverse order. Apply every operation to the whole of both sides, then substitute the rearranged expression back into the original relationship as a check.
Do not move a term by changing its sign without performing an inverse operation on both sides. Keep brackets when an operation applies to a complete sum or difference.
Substitution replaces each symbol with its measured value only after the values have been expressed in units consistent with the equation.
| Step | Action |
|---|---|
| 1 | write the equation and identify every symbol |
| 2 | convert values to one compatible unit system |
| 3 | substitute with brackets around negative or compound values |
| 4 | calculate while retaining meaningful intermediate digits |
| 5 | give the answer with its derived unit and appropriate precision |
c=\frac{n}{V}=\frac{0.250,\mathrm{mol}}{0.500,\mathrm{dm^3}}=0.500,\mathrm{mol,dm^{-3}}
Units behave algebraically: dividing mol by dm3 gives moldm−3. Convert 500cm3 to 0.500dm3 before using a concentration equation expressed per dm3.
A numerically correct substitution with incompatible units is not a valid physical result. Do not append a memorised unit without deriving it from the equation.
Solving an equation finds the value that makes both sides equal by preserving the balance at every step.
| Step | Example for 3x+5=20 |
|---|---|
| simplify each side if needed | already simplified |
| undo addition or subtraction | 3x=15 |
| undo multiplication or division | x=5 |
| check in the original equation | 3(5)+5=20 |
Collect like terms before isolating the unknown. If the unknown appears in a denominator, multiply both sides by the denominator where it is non-zero; if brackets occur, expand them or undo their outer operation consistently.
An equation may have a restriction, such as a denominator not being zero. Do not accept a value until substitution confirms that it satisfies the original equation and its physical context.