For centuries, algebra was written in words. In the 16th and 17th centuries, mathematicians such as François Viète began to use letters systematically for both unknowns and known parameters. That made formulas — general relationships such as \(v = d/t\) — easy to rearrange without repeating long verbal rules.
Rational expressions (fractions built from polynomials) appear whenever you divide one polynomial by another. Scientists from Galileo onward used ratios and proportionalities in motion; later, Newton’s laws and electrical laws (e.g. \(V = IR\)) relied on changing the subject of a formula to solve for the quantity you can measure.
Today, algebraic fractions show up in simplifying models, in calculus limits, and whenever a relationship is written as one quantity divided by another.
If \(c \neq 0\), then \(\dfrac{ac}{bc} = \dfrac{a}{b}\) because multiplying numerator and denominator by the same non-zero number does not change the value of a numeric fraction; the same idea holds when \(c\) is a polynomial factor shared by top and bottom.
You cannot cancel in \(\dfrac{a+c}{b+c}\) in general — that would mean, for example, \(\dfrac{1+2}{3+2} = \dfrac{1}{3}\), which is false. Only factors, not terms, may be cancelled.
A formula like \(v = d/t\) builds \(v\) by first dividing \(d\) by \(t\). To recover \(t\), undo division by multiplying: \(vt = d\), then undo multiplication by \(v\) (when \(v \neq 0\)) to get \(t = d/v\). Each step reverses one operation that was applied to the subject you want.
Pause and reason before continuing. Discuss with a partner or write your work in a notebook.
Checkpoint
Simplify \(\dfrac{x^2 - 4}{x + 2}\). What value of \(x\) must be excluded?
Checkpoint
Can you cancel the \(x\) in \(\dfrac{x + 1}{x}\)? Explain briefly.
Checkpoint
Write \(\dfrac{3}{x} - \dfrac{1}{2x}\) as a single fraction in simplest form.
Checkpoint
Make \(d\) the subject of \(v = \dfrac{d}{t}\). State any conditions on \(t\) and \(v\).
Checkpoint
A wire obeys \(V = IR\). If voltage and resistance are known, which formula gives current?
Checkpoint
Simplify \(\dfrac{1}{x} + \dfrac{1}{x+1}\). What is the common denominator?
For constant speed, distance \(d\), speed \(v\), and time \(t\) are related by \(v = d/t\). If a train travels \(240\) km in \(3\) h, then:
\[ \begin{aligned} v &= \frac{240}{3} \\ &= 80\ \text{km/h} \end{aligned} \]
If you know \(v\) and \(t\) but need distance, rearrange to \(d = vt\).
Density \(\rho\) (rho) is mass per unit volume: \(\rho = m/V\). A block of mass \(2.7\) kg with volume \(0.001\) m³ has:
\[ \begin{aligned} \rho &= \frac{2.7}{0.001} \\ &= 2700\ \text{kg/m}^3 \end{aligned} \]
To find mass from \(\rho\) and \(V\), use \(m = \rho V\).
Pressure \(P = F/A\) links force \(F\) and area \(A\). Hydraulic and weather models often solve for \(F = PA\). In a resistor, \(V = IR\) gives voltage from current and resistance; rearrange to \(I = V/R\) when you know voltage and resistance.
Work each question, then open the solution to check your reasoning. Difficulty increases through the sets.
Answer: \(2\), with \(x \neq 0\).
Steps: Cancel the common factor \(3x\).
Answer: \(x\).
Explanation: Cancel one factor \(x\) from numerator and denominator.
Answer: \(d = vt\) (\(t \neq 0\) when rearranging from the original).
Steps: Multiply both sides by \(t\).
Answer: \(R = V/I\) (\(I \neq 0\)).
Steps: Divide both sides by \(I\).
Answer: \(x - 3\), with \(x \neq -3\).
Steps: \(x^2 - 9 = (x-3)(x+3)\); cancel \((x+3)\).
Answer: \(\dfrac{3}{x}\), \(x \neq 0\).
Explanation: Same denominator already.
Answer: \(\dfrac{x-1}{x}\), \(x \neq 0,\; x \neq -1\).
Steps: Factorise; cancel \((x+1)\).
Answer: \(m = \rho V\) (\(V \neq 0\) in the original).
Steps: Multiply both sides by \(V\).
Answer: \(\dfrac{3x+1}{x(x+1)}\), \(x \neq 0,\; x \neq -1\).
Steps: LCD \(x(x+1)\); numerator \( (x+1) + 2x = 3x+1\).
Answer: \(\dfrac{2}{x}\), \(x \neq 0\).
Steps: Multiply; cancel a factor \(x\).
Answer: \(\dfrac{x-1}{(x-2)(x+2)}\), \(x \neq 2,\; x \neq -2\).
Steps: Write \(x^2-4 = (x-2)(x+2)\); common denominator \((x-2)(x+2)\).
Answer: \(2\), with \(x \neq -1,\; x \neq 1\).
Steps: Factorise \(x^2-1\); multiply by reciprocal; cancel \((x-1)\) and \((x+1)\).
Answer:
\[ \begin{aligned} t &= \frac{d}{v} \\ &= \frac{150}{75} \\ &= 2\ \text{h} \end{aligned} \]
Steps: \(vt = d \Rightarrow t = d/v\); substitute values.
Answer: \(2\), \(x \neq 2,\; x \neq -2\).
Steps: Factorise numerator \(2(x^2-4)\) and denominator \((x-2)(x+2)\); cancel \((x-2)(x+2)\).
Answer:
\[ \begin{aligned} P &= \frac{120}{0.05} \\ &= 2400\ \text{Pa} \end{aligned} \]
\[ \begin{aligned} F &= PA \\ &= 2400 \times 0.05 \\ &= 120\ \text{N} \end{aligned} \]
Explanation: Rearranging checks that the formula is consistent.