⚡ Quick Memory Points
🔑 CROSS-MULTIPLY
The Golden Rule
When two fractions are equal, cross-multiply to eliminate denominators instantly.
a/b = c/d → ad = bc
🚫 EXTRANEOUS
Always Check!
After solving, plug your answer back in. If the denominator becomes 0, that solution is extraneous (rejected).
denominator ≠ 0 ← CHECK THIS!
🔗 LCD METHOD
Least Common Denom.
Multiply every term by the LCD to clear all fractions at once. Works like magic.
× LCD → no more fractions
📐 PROPORTION
Equal Ratios
A proportion is two equal ratios. Cross-multiply only works when it's fraction = fraction (nothing else on either side).
Only: a/b = c/d form
♻️ FACTOR FIRST
Simplify Before Solving
Always factor numerators and denominators before doing anything. Cancel common factors — but remember the restriction!
Factor → Cancel → Solve
⚖️ BALANCE
Same Operation Both Sides
Whatever you do to one side of the equation, do to the other. Multiply, add, subtract — keep it balanced.
Equation stays true if balanced
Part 1 — Foundations
Q1–7
1
Solve for \(x\): \(\dfrac{x}{4} = \dfrac{3}{2}\)
💡 Hint: Use cross-multiply: multiply diagonally across the equal sign.
📖 Solution
Cross-multiply: \(x \cdot 2 = 4 \cdot 3\) → \(2x = 12\) → \(x = 6\). ✓ Check: \(\frac{6}{4} = \frac{3}{2}\) ✓
2
What value of \(x\) makes \(\dfrac{2}{x} = \dfrac{1}{5}\)?
💡 Hint: Cross-multiply: \(2 \times 5 = 1 \times x\).
📖 Solution
Cross-multiply: \(2 \times 5 = 1 \times x\) → \(x = 10\). Check: \(\frac{2}{10} = \frac{1}{5}\) ✓
3
Solve: \(\dfrac{x+1}{3} = \dfrac{x-1}{2}\)
💡 Hint: Cross-multiply first, then distribute and isolate \(x\).
📖 Solution
Cross-multiply: \(2(x+1) = 3(x-1)\) → \(2x+2 = 3x-3\) → \(-x = -5\) → \(x = 5\). ✓
4
Which value of \(x\) must be excluded from the domain of \(\dfrac{x+2}{x-3}\)?
💡 Hint: The denominator can never equal zero. Set denominator = 0 and solve.
📖 Solution
Set denominator = 0: \(x - 3 = 0\) → \(x = 3\). So \(x = 3\) is excluded (undefined). The numerator's zero (\(x = -2\)) is fine — it just makes the fraction equal to 0.
5
Simplify: \(\dfrac{2x}{6x^2}\) (assume \(x \neq 0\))
💡 Hint: Factor and cancel. \(2x\) divides into \(6x^2\) — how many times?
📖 Solution
\(\dfrac{2x}{6x^2} = \dfrac{2x}{6x \cdot x} = \dfrac{1}{3x}\). Cancel the \(2x\) from top and bottom.
6
Solve for \(x\): \(\dfrac{3}{x} + \dfrac{1}{2} = \dfrac{5}{2}\)
💡 Hint: First subtract \(\frac{1}{2}\) from both sides, then cross-multiply to solve for \(x\).
📖 Solution
Subtract \(\frac{1}{2}\) from both sides: \(\frac{3}{x} = \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2\). Cross-multiply: \(3 = 2x\) → \(x = \frac{3}{2}\). Check: \(\frac{3}{3/2} + \frac{1}{2} = 2 + \frac{1}{2} = \frac{5}{2}\) ✓. Key: Isolate the fraction with \(x\) first, then cross-multiply!
7
Which is the LCD for \(\dfrac{1}{x}\) and \(\dfrac{1}{x+2}\)?
💡 Hint: LCD = the product of two different factors when they share no common factor.
📖 Solution
Since \(x\) and \(x+2\) share no common factor, LCD = \(x(x+2)\). This is the smallest expression that both denominators divide into evenly.
Part 2 — Core Problems
Q8–14
8
Solve: \(\dfrac{5}{x-2} = \dfrac{3}{x+1}\)
💡 Hint: Cross-multiply → distribute → collect \(x\) terms on one side.
📖 Solution
Cross-multiply: \(5(x+1) = 3(x-2)\) → \(5x+5 = 3x-6\) → \(2x = -11\) → \(x = -\frac{11}{2}\). Check denominators: \(x-2 = -\frac{15}{2} \neq 0\) ✓, \(x+1 = -\frac{9}{2} \neq 0\) ✓
9
Solve: \(\dfrac{x}{x-3} = \dfrac{3}{x-3}\)
💡 Hint: The denominators are the same — this looks easy, but be careful! Check your answer!
📖 Extraneous Solution Trap!
Since denominators are equal: \(x = 3\). But wait — plug \(x=3\) back in: denominator = \(3-3 = 0\)! ❌ Division by zero is undefined. So \(x = 3\) is extraneous and must be rejected. Answer: No solution. This is the #1 most common mistake — always check your answer!
10
Add and simplify: \(\dfrac{1}{x} + \dfrac{1}{x+1}\)
💡 Hint: Find LCD = \(x(x+1)\), then rewrite each fraction with that denominator.
📖 Solution
LCD = \(x(x+1)\). \(\dfrac{x+1}{x(x+1)} + \dfrac{x}{x(x+1)} = \dfrac{x+1+x}{x(x+1)} = \dfrac{2x+1}{x(x+1)}\). Can't simplify further.
11
Solve: \(\dfrac{2}{x+1} + \dfrac{3}{x-1} = \dfrac{5}{x^2-1}\)
💡 Hint: Notice that \(x^2-1 = (x+1)(x-1)\). That's your LCD! Multiply everything by \((x+1)(x-1)\).
📖 Solution
LCD = \((x+1)(x-1)\). Multiply through: \(2(x-1) + 3(x+1) = 5\) → \(2x-2+3x+3 = 5\) → \(5x+1 = 5\) → \(5x = 4\) → \(x = \frac{4}{5}\). Check: \(x = \frac{4}{5}\) makes no denominator zero (not ±1) ✓. Key: Recognize \(x^2-1 = (x+1)(x-1)\) to find the LCD instantly!
12
Simplify: \(\dfrac{x^2 - 4}{x^2 - 2x}\)
💡 Hint: Factor top: difference of squares. Factor bottom: pull out \(x\).
📖 Solution
\(\dfrac{x^2-4}{x^2-2x} = \dfrac{(x+2)(x-2)}{x(x-2)} = \dfrac{x+2}{x}\). Cancel \((x-2)\), but note restrictions: \(x \neq 0\) and \(x \neq 2\). Key: Always state domain restrictions after canceling!
13
Solve: \(\dfrac{x}{x+2} = 1\)
💡 Hint: If \(\dfrac{x}{x+2} = \dfrac{1}{1}\), can you cross-multiply? Or multiply both sides by \((x+2)\)?
📖 Tricky One!
Multiply both sides by \((x+2)\): \(x = 1\cdot(x+2)\) → \(x = x + 2\) → \(0 = 2\). This is a contradiction! No value of \(x\) satisfies this. Answer: No solution. Intuitively, \(\frac{x}{x+2}\) can never equal 1 because the numerator is always 2 less than the denominator.
14
Solve: \(\dfrac{1}{x-1} - \dfrac{1}{x+1} = \dfrac{2}{x^2-1}\)
💡 Hint: \(x^2-1 = (x-1)(x+1)\) — recognize the pattern! LCD = \((x-1)(x+1)\).
📖 Solution — Identity!
Multiply by LCD = \((x-1)(x+1)\): \((x+1) - (x-1) = 2\) → \(x+1-x+1 = 2\) → \(2 = 2\). This is always true! It's an identity — the equation holds for all \(x\), except where denominators = 0 (i.e., \(x \neq \pm 1\)).
Part 3 — Tricky Traps
Q15–20
15
Solve: \(\dfrac{2x}{x-1} - 3 = \dfrac{x+1}{x-1}\)
💡 Hint: Multiply both sides by \((x-1)\). Don't forget to distribute the 3!
📖 Extraneous Solution Trap!
Multiply both sides by \((x-1)\): \(2x - 3(x-1) = x+1\) → \(2x - 3x + 3 = x + 1\) → \(-x + 3 = x + 1\) → \(2 = 2x\) → \(x = 1\). But check: \(x = 1\) makes the denominator \(x-1 = 0\)! ❌ Extraneous! So there is No solution. The algebra gave us an answer, but it's invalid — this is why you must ALWAYS check your answer by substituting back!
16
If \(\dfrac{a}{b} = \dfrac{3}{5}\), what is \(\dfrac{a+b}{b}\)?
💡 Hint: Split the fraction: \(\dfrac{a+b}{b} = \dfrac{a}{b} + \dfrac{b}{b}\). Then substitute!
📖 Solution — Elegant Trick
\(\dfrac{a+b}{b} = \dfrac{a}{b} + \dfrac{b}{b} = \dfrac{3}{5} + 1 = \dfrac{3}{5} + \dfrac{5}{5} = \dfrac{8}{5}\). No need to find actual values of \(a\) and \(b\)!
17
Simplify: \(\dfrac{x^2 - x - 6}{x^2 - 9}\)
💡 Hint: Factor both. Numerator: find two numbers that multiply to −6 and add to −1. Denominator: difference of squares!
📖 Solution
Numerator: \(x^2-x-6 = (x-3)(x+2)\). Denominator: \(x^2-9 = (x-3)(x+3)\). Cancel \((x-3)\): \(\dfrac{x+2}{x+3}\) (where \(x \neq 3, -3\)). Common trap: confusing the signs when factoring the numerator.
18
Solve: \(\dfrac{1}{x-2} + \dfrac{1}{x+2} = \dfrac{4}{x^2-4}\)
💡 Hint: \(x^2-4 = (x-2)(x+2)\). After multiplying by LCD, check if your answer makes any denominator = 0!
📖 Solution
LCD = \((x-2)(x+2)\). Multiply: \((x+2) + (x-2) = 4\) → \(2x = 4\) → \(x = 2\). But \(x=2\) makes \(x-2 = 0\)! Extraneous! So the answer is No solution.
19
If \(\dfrac{p}{q} = \dfrac{2}{3}\), which of the following must be true?
💡 Hint: Try \(p=2, q=3\) and test each option. A proportion gives us a specific ratio — not specific values.
📖 Solution
Given \(\frac{p}{q} = \frac{2}{3}\), i.e., \(3p = 2q\). Test C: \(\frac{3p}{2q} = \frac{3p}{2q}\). Since \(3p = 2q\), we get \(\frac{2q}{2q} = 1\) ✓. A is wrong (\(p=4, q=6\) also works). B is wrong (\(p+q = 4+6=10 \neq 5\)). D is wrong (\(4-6=-2\neq -1\)). Only C must always be true.
20
Solve: \(\dfrac{x^2 - 1}{x - 1} = x + 2\)
💡 Hint: Factor the numerator first. You can cancel — but what's the restriction? After simplifying, does the resulting equation have a valid solution?
📖 Final Boss Solution
Factor: \(\dfrac{(x+1)(x-1)}{x-1} = x+2\). For \(x \neq 1\), simplify: \(x+1 = x+2\) → \(1 = 2\). Contradiction! No solution. The simplification reveals an impossibility. And \(x=1\) is excluded anyway (denominator = 0). The trap: students cancel and then forget the equation becomes a contradiction. Always finish solving after you cancel!
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