Algebraic Fractions: Cross-Multiply

What does “cross-multiply” mean?

When two fractions are equal, \[ \frac{A}{B} = \frac{C}{D}, \] and the denominators are not zero (\(B\neq 0\), \(D\neq 0\)), we can multiply both sides by the common denominator \(BD\) to remove the fractions: \[ BD\cdot\frac{A}{B} = BD\cdot\frac{C}{D} \quad\Longrightarrow\quad AD = BC. \] This step is often called cross-multiplying because the numerators and denominators “cross” (top left with bottom right, top right with bottom left).

Why it works (one line reason)

It’s just multiplying both sides by the same non-zero quantity \(BD\). Because \(BD\neq 0\), the equality is preserved. After cancellation we are left with \(AD=BC\), a fraction-free equation that is usually easier to solve.

Basic examples

• \(\dfrac{x}{5}=\dfrac{3}{4}\). Cross-multiply: \(4x=15\Rightarrow x=\tfrac{15}{4}=3.75\).
• \(\dfrac{2}{3}=\dfrac{x}{9}\). Cross-multiply: \(3x=18\Rightarrow x=6\).
• \(\dfrac{x-1}{4}=\dfrac{5}{2}\). Cross-multiply: \(2(x-1)=20\Rightarrow x-1=10\Rightarrow x=11\).

With algebra inside the fractions

If the unknown appears in the numerator or denominator, the method is the same, but we must remember the domain restrictions (values that make a denominator zero are not allowed).

• \(\dfrac{x+2}{x-1}=\dfrac{3}{2}\), with \(x\neq 1\).
  Cross-multiply: \(2(x+2)=3(x-1)\Rightarrow 2x+4=3x-3\Rightarrow x=7\). Check \(x\neq 1\) ✓.
• \(\dfrac{5}{x+1}=\dfrac{10}{3}\), with \(x\neq -1\).
  Cross-multiply: \(3\cdot 5 = 10(x+1)\Rightarrow 15=10x+10\Rightarrow x=\tfrac{1}{2}\).
• \(\dfrac{2x-3}{x+4}=\dfrac{1}{2}\), with \(x\neq -4\).
  Cross-multiply: \(2(2x-3)=x+4\Rightarrow 4x-6=x+4\Rightarrow 3x=10\Rightarrow x=\tfrac{10}{3}\).

When not to cross-multiply

• Not an equation of one fraction = one fraction. For example \( \dfrac{x}{3}+\dfrac{1}{2}=5 \) is a sum of fractions; you should use a common denominator (here \(6\)) rather than cross-multiplying.
• Different structure. Something like \( \dfrac{x}{x+1}=3 \) can be solved by multiplying both sides by \(x+1\) directly; there is no “other” fraction to cross with.
• Hidden zero denominators. Always state \(B\neq 0\) and \(D\neq 0\) before you cross-multiply. Any solution that makes a denominator zero must be rejected at the end.

Step-by-step method (safe recipe)

1. Write domain restrictions: list values that make any denominator zero (e.g. \(x\neq -1,\,2\)).
2. Cross-multiply: if \(\dfrac{A}{B}=\dfrac{C}{D}\) then set \(AD=BC\).
3. Expand and simplify: remove brackets, collect like terms.
4. Solve the resulting linear (or sometimes quadratic) equation.
5. Check against the domain: reject any solution that makes a denominator zero.

Proportion and ratio applications

Cross-multiplying is the algebra behind “proportion triangles” and ratio tables. If two quantities are proportional, \(\dfrac{x}{y}=\dfrac{k_1}{k_2}\), then \(k_2x=k_1y\).

• Recipe scaling: If \( \dfrac{\text{sugar}}{\text{flour}}=\dfrac{3}{5} \) and flour \(=400\text{ g}\), then \( \dfrac{s}{400}=\dfrac{3}{5} \Rightarrow 5s=1200 \Rightarrow s=240\text{ g}\).
• Speed (\(s=\dfrac{d}{t}\)): If \( \dfrac{d_1}{t_1}=\dfrac{d_2}{t_2}\) at constant speed, then \( d_1 t_2 = d_2 t_1\). This is cross-multiplying the equality of speeds.
• Similar triangles: Corresponding sides are in proportion, so lengths satisfy a fraction equality that can be solved by cross-multiplying.

Typical GCSE patterns

1) One linear term each side

\[ \frac{ax+b}{c}=\frac{d}{e} \quad\Longrightarrow\quad e(ax+b)=cd \quad\Longrightarrow\quad ax+b=\frac{cd}{e}. \] Then solve \(ax+b=\ldots\) as usual.

2) Unknown in numerators and denominators

\[ \frac{ax+b}{cx+d}=\frac{p}{q}, \qquad cx+d\neq 0. \] Cross-multiply: \( q(ax+b)=p(cx+d) \). Expand and solve. Remember to exclude \(x\) that make \(cx+d=0\).

Example: \( \dfrac{x+5}{2x-3}=\dfrac{4}{7} \Rightarrow 7(x+5)=4(2x-3) \Rightarrow 7x+35=8x-12 \Rightarrow x=47.\) Domain check: \(2x-3\neq 0\) gives \(x\neq \tfrac{3}{2}\); our solution \(47\) is valid.

3) Creating a quadratic

Sometimes cross-multiplying leads to a quadratic equation.

Example: \( \dfrac{x}{x-2}=\dfrac{3}{x} \) with \(x\neq 0,2\). Cross-multiply: \(x\cdot x = 3(x-2)\Rightarrow x^2=3x-6\Rightarrow x^2-3x+6=0\). This quadratic has discriminant \(9-24<0\), so there is no real solution. (That’s fine—sometimes proportions have no real solution once the domain is enforced.)

Common errors (and fixes)

• Forgetting domain restrictions. Always state forbidden values first, and check at the end. If a candidate solution makes a denominator zero, it must be rejected.
• Cross-multiplying when there isn’t a single fraction on each side. Convert sums/differences to a single fraction using an LCM; then proceed.
• Dropping brackets. When you write \(e(ax+b)=\ldots\), multiply everything inside: \(eax+eb\).
• Sign slips. Keep equals aligned. After cross-multiplying, work line-by-line and avoid skipping steps.

Worked examples in full sentences

1. Solve: \(\dfrac{3x+1}{5}=\dfrac{7}{10}\).
   Cross-multiply: \(10(3x+1)=5\cdot 7\Rightarrow 30x+10=35\Rightarrow 30x=25\Rightarrow x=\tfrac{5}{6}\).
2. Solve (state domain): \(\dfrac{x-4}{x+2}=\dfrac{2}{3}\).
   Domain: \(x\neq -2\). Cross-multiply: \(3(x-4)=2(x+2)\Rightarrow 3x-12=2x+4\Rightarrow x=16\) (valid).
3. Similar-triangles side: If \(\dfrac{AB}{DE}=\dfrac{BC}{EF}\) and \(AB=7\), \(DE=5\), \(EF=12\), find \(BC\).
   \(\dfrac{7}{5}=\dfrac{BC}{12}\Rightarrow 5\cdot BC=84\Rightarrow BC=16.8\).
4. Unit conversion ratio: \(\dfrac{\text{miles}}{\text{hours}}=\dfrac{150}{3}=\dfrac{50}{1}\). If time is \(2.4\) hours at the same speed, distance \(d\) satisfies \(\dfrac{d}{2.4}=\dfrac{50}{1}\Rightarrow d=120\) miles.

Cross-multiplying in inequalities (careful!)

For equations cross-multiplying is always safe with \(BD\neq 0\). For inequalities (\(\lt,\le,\gt,\ge\)) you must know the sign of what you multiply by, because multiplying by a negative reverses the inequality sign.

• If you know both denominators are positive (e.g. they are lengths or squares), you can multiply by the positive \(BD\) safely.
• If signs are unknown, rearrange to put everything on one side or solve by case-splitting on the sign of the denominator.

Example: For \(x>0\) the inequality \(\dfrac{1}{x}\le \dfrac{2}{3}\) can be cross-multiplied (since \(x\) and \(3\) are positive): \(3\le 2x\Rightarrow x\ge \tfrac{3}{2}\). If \(x\) were not known to be positive, you would solve by considering \(x>0\) and \(x<0\) separately.

Strategy for exam questions

1. Check the structure: do you truly have one fraction equal to one fraction?
2. Write domain restrictions: list values that make denominators zero.
3. Cross-multiply: set \(AD=BC\).
4. Expand/collect: remove brackets and bring like terms together.
5. Solve the resulting equation.
6. Validate: discard any value that breaks the domain; if none remain, state “no real solution.”

Mini practice (try mentally)

1. \(\dfrac{x}{7}=\dfrac{5}{14}\)
2. \(\dfrac{x-3}{x+1}=\dfrac{4}{5}\) (state the domain)
3. \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
4. \(\dfrac{p}{p-2}=\dfrac{3}{4}\) (find forbidden values and solve)
5. The ratio \(a:b=5:8\) and \(b=56\). Find \(a\).

Summary

Cross-multiplying is a reliable way to clear fractions in a proportion: from \( \dfrac{A}{B}=\dfrac{C}{D} \) (with \(B,D\neq 0\)) we get \(AD=BC\). It turns many algebraic fraction equations into straightforward linear (or occasionally quadratic) equations. Always record domain restrictions, expand carefully with brackets, and validate solutions at the end. For inequalities, be mindful of signs before multiplying. With these habits, proportion and algebraic-fraction questions become fast and error-free.