Math Vale

⚔️ Algebra Kingdom · Algebra

Equations with Fractions

Clear fractions from an equation by multiplying through by the least common denominator, then solve as usual.

In short

  • Multiplying every term on both sides by the least common denominator turns a fraction equation into an ordinary one.
  • The LCD is the least common multiple of the denominators; any common multiple works, the least one just keeps numbers small.
  • To undo multiplication by a fraction, multiply by its reciprocal — but clear any constant term first.
  • Two equal fractions can be cross-multiplied, and if the unknown is in a denominator you must exclude the value that makes it zero.

Make the fractions leave

An equation with fractions is not a new kind of equation. It is an ordinary equation wearing an awkward costume, and there is a move that takes the costume off in one go.

Multiply every term on both sides by the least common denominator. After that step there are no fractions left and you are back to the equations you already know how to solve.

For x/3 + 2 = 7 the only denominator is 3, so multiply everything by 3: x + 6 = 21, and x = 15. Check in the original: 15/3 + 2 = 5 + 2 = 7.

Finding the least common denominator

The least common denominator (LCD) of the fractions in an equation is the smallest number that every denominator divides into exactly — the least common multiple of the denominators.

  • Denominators 2 and 3: the LCD is 6.
  • Denominators 4 and 6: the LCD is 12, not 24. Smaller keeps the numbers friendlier.
  • Denominators 3 and 6: the LCD is 6, because 3 already divides 6.

Any common multiple works and gives the same solution; the least one just keeps the arithmetic small. If you are unsure, multiplying all the denominators together is always a safe common multiple.

Every term, both sides

This is where nearly all the marks are lost. Multiplying by the LCD is a balance move, so it must reach every term on both sides — the fractions, the plain numbers, and the right-hand side.

Take x/2 + x/3 = 10 with LCD 6:

  • 6 x (x/2) = 3x
  • 6 x (x/3) = 2x
  • 6 x 10 = 60

so 3x + 2x = 60, then 5x = 60 and x = 12. Check: 12/2 + 12/3 = 6 + 4 = 10.

Notice what did not happen: the denominators were never added. x/2 + x/3 is not x/5. Fractions with different denominators are combined by rewriting them over a common denominator, never by adding the bottoms.

Fractional coefficients and the reciprocal

When the fraction multiplies the unknown, as in (2/3)x = 8, there is a second route that is often quicker: multiply both sides by the reciprocal.

The reciprocal of 2/3 is 3/2, and (3/2) x (2/3) = 1, so multiplying by 3/2 frees x: x = 8 x 3/2 = 12.

Both routes agree, because multiplying by 3/2 is the same as multiplying by 3 and dividing by 2. Use whichever you find easier — but if there is also a constant, as in (2/3)x + 5 = 13, deal with the constant first: (2/3)x = 8, then apply the reciprocal.

Proportions and forbidden values

Two equal fractions form a proportion, and multiplying both sides by both denominators gives the familiar cross-multiplication: from a/b = c/d you get ad = bc. It is not a separate rule, just the LCD move written quickly.

So 4/5 = x/15 becomes 4 x 15 = 5x, giving 5x = 60 and x = 12.

When the unknown sits in a denominator, one extra question comes first: which values are not allowed? In 6/x = 3/8, x cannot be 0, because nothing may be divided by zero. Cross-multiplying gives 3x = 48 and x = 16, which is not 0, so it is a genuine solution. Always check your answer against the excluded values before you write it down.

Worked examples

Example 1

Solve for x: x/4 + x/6 = 5

  1. The denominators are 4 and 6, so the least common denominator is 12.
  2. Multiply every term on both sides by 12: 12 x (x/4) + 12 x (x/6) = 12 x 5.
  3. That gives 3x + 2x = 60, with no fractions left.
  4. Combine like terms: 5x = 60, so x = 12.
  5. Check in the original: 12/4 + 12/6 = 3 + 2 = 5. Correct.

Example 2

A quartermaster records that 3/5 of the store of lanterns has been issued, which came to 42 lanterns. How many lanterns were in the store?

  1. Let n be the number of lanterns in the store at the start.
  2. "3/5 of n" is (3/5)n, and the story says that equals 42: (3/5)n = 42.
  3. Multiply both sides by the reciprocal 5/3: n = 42 x 5/3.
  4. n = 210/3 = 70 lanterns.
  5. Check: 3/5 of 70 is 42, exactly as recorded. Correct.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Solve for x: (x + 3)/5 = 4

Answer: 17

  1. (x + 3)/5 = 4
  2. x + 3 = 5 x 4 = 20 (multiply both sides by 5)
  3. x = 20 - 3
  4. x = 17
  5. Check: (17 + 3)/5 = 4. Correct.

Problem 2

Difficulty 3 of 5

Solve for w: (2/3)w = 8

Answer: 12

  1. (2/3)w = 8
  2. (3/2) x (2/3)w = (3/2) x 8 (multiply both sides by the reciprocal)
  3. w = 24/2 = 12
  4. Check: (2/3) x 12 = 8. Correct.

Problem 3

Difficulty 4 of 5

Solve for w: w/3 + w/8 - 3 = -1025/8

Answer: -273

  1. w/3 + w/8 - 3 = -1025/8
  2. The least common denominator of 3 and 8 is 24.
  3. Multiply every term by 24: 8w + 3w - 72 = -3075
  4. 11w = -3003 (combine like terms and clear the constant)
  5. w = -273. Check: (-273)/3 + (-273)/8 - 3 = -1025/8.

Common mistakes

  • Multiplying only the fractions by the LCD and leaving the whole-number terms untouched.
  • Multiplying one side of the equation by the LCD and forgetting the other side entirely.
  • Adding denominators: treating x/2 + x/3 as x/5.
  • Multiplying by the coefficient instead of its reciprocal, so (2/3)x = 8 gives 16/3 rather than 12.

What you should be able to do

  • Multiply both sides by the least common denominator to clear fractions.
  • Solve equations with fractional coefficients and constants.
  • Solve simple proportions as rational equations.
  • Check solutions and exclude values that make a denominator zero.

Where this fits in the curriculum

Common Core

  • 8.EE.C.7.B

    Grade 8 — Solve linear equations with rational number coefficients, including those requiring the distributive property.

  • HSA-REI.A.1

    High school — Explain each step in solving an equation as following from the equality of the previous step.

  • HSA-REI.A.2

    High school — Solve simple rational equations, and show how extraneous solutions can arise.

    Clearing denominators with the LCD is a technique rather than a standard of its own; HSA-REI.A.2 is the standard that owns "check for values that make a denominator zero".

Ontario

  • MTH1W.C1.5

    Grade 9 de-streamed — Create and solve equations for various contexts, and verify their solutions.

  • MTH1W.B3.4

    Grade 9 de-streamed — Solve problems involving operations with positive and negative fractions and mixed numbers, including in linear relations.

    Ontario has no expectation for "clear the denominators with the LCD"; B3.4 is the nearest, framed as fraction operations rather than as an equation-solving technique, so this pairing is a judgement call.

SAT

  • Heart of Algebra

    Linear equations with rational coefficients.

  • Passport to Advanced Math

    Rational equations and expressions, including extraneous solutions.

    This skill straddles two domains: clearing denominators in a linear equation is Heart of Algebra, while a genuine rational equation is Passport to Advanced Math.

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