--- title: "Rearranging formulae when the subject appears twice" description: "Change the subject of a formula where the new subject appears more than once: clear fractions, collect its terms, factorise it out and divide by a nonzero bracket. Check excluded cases separately." canonical: https://lightmysky.com/learn/mathematics/rearranging-formulae-when-the-subject-appears-twice-mt_Velmpnxfcj source: https://lightmysky.com/learn/mathematics/rearranging-formulae-when-the-subject-appears-twice-mt_Velmpnxfcj.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Rearranging formulae when the subject appears twice Change the subject of a formula where the new subject appears more than once: clear fractions, collect its terms, factorise it out and divide by a nonzero bracket. Check excluded cases separately. Subject: Mathematics · Area: Algebra · Ages 15 to 16 Page: https://lightmysky.com/learn/mathematics/rearranging-formulae-when-the-subject-appears-twice-mt_Velmpnxfcj ## Ready when they can - Multiply through to clear a fraction before collecting terms - Gather every term containing the new subject on one side - Factorise the subject out of the collected terms and divide by a nonzero bracket, checking zero-bracket cases separately ## Lesson: Rearranging formulae when the subject appears twice You have factorised algebraic fractions and recorded excluded inputs. To make a letter the subject, write it alone on one side. If it appears in several terms, multiply every term on both sides by a common denominator to clear fractions. Keep the original denominator restrictions. Then collect the wanted letter's terms on one side, factorise it out and divide by the remaining bracket, provided that bracket is nonzero. *(drawing: Record original restrictions before clearing fractions.)* Collecting gives a common factor: xy - 2x = B becomes x(y - 2) = B. Divide by y - 2 to isolate x. Dividing by x instead could discard a solution with x = 0. A quotient is undefined when its denominator is zero. For x = (2y + 1)/(y - 4), solve y - 4 = 0 to find the excluded value y = 4. Check the equation before division separately at any value excluded by a new divisor. **Example.** Make x the subject of y = (2x + 5)/(x - 3), with x != 3. Multiply by x - 3: y(x - 3) = 2x + 5. Expand: xy - 3y = 2x + 5. Subtract 2x and add 3y on both sides: xy - 2x = 3y + 5. Factorise: x(y - 2) = 3y + 5. For y != 2, divide by the whole bracket: x = (3y + 5)/(y - 2). If y = 2, the collected equation reads 0 = 11, so there is no solution. **Example.** For A = 3x + 2xy, factorise: A = x(3 + 2y). When 3 + 2y != 0, x = A/(3 + 2y). Subtract any term without x before factorising. Subtract x terms on the other side to collect them too. Substitute known values into the final expression to calculate x. If 3 + 2y = 0, use A = 0x instead: A != 0 gives no solution; A = 0 allows every real x. **Example.** Make x the subject of R = 1/x + 1/2, with x != 0. The original has one x, but clearing fractions creates two x terms. Multiply every term by 2x: 2xR = 2 + x. Subtract x: 2xR - x = 2. Factorise: x(2R - 1) = 2. For R != 1/2, divide: x = 2/(2R - 1). At R = 1/2, the collected equation is 0 = 2, so there is no solution. Check a proposed value in the original formula, including its denominator. In y = (2x + 5)/(x - 3), setting y = 3 gives x = 14 from the rearranged expression. Substitution returns (2(14) + 5)/(14 - 3) = 33/11 = 3. A different result would disprove the proposed solution. You may also substitute a whole rearranged expression and simplify to check it algebraically. **Example.** To change back from x = (3y + 5)/(y - 2), make y the subject. Multiply by y - 2: xy - 2x = 3y + 5. Subtract 3y and add 2x: xy - 3y = 2x + 5. Factorise y: y(x - 3) = 2x + 5. Divide to get y = (2x + 5)/(x - 3), with x != 3 and y != 2. To identify constants in y = (ax + b)/(x - c), compare coefficients: a = 2, b = 5 and c = 3. **Example.** Brackets may hide x terms. In (t - 2)(x + 3) = 4x, expand: x(t - 2) + 3(t - 2) = 4x. Subtract 4x and 3(t - 2) on both sides: x(t - 2) - 4x = -3(t - 2). Factorise: x(t - 6) = -3(t - 2). For t != 6, divide: x = -3(t - 2)/(t - 6) = 3(t - 2)/(6 - t). Multiplying the numerator and denominator by -1 changes both signs and keeps the quotient equal. **Recap.** Clear fractions, collect the wanted letter's terms, factorise and divide by a nonzero factor. Keep original restrictions and check zero-factor cases before division. *This lesson, its practice questions and its flashcards were written for this dataset.* ## Practice 28 questions on this page, each with its working shown. ## Needs first - [Simplifying algebraic fractions](https://lightmysky.com/learn/mathematics/simplifying-algebraic-fractions-mt_scDNSoz3eW) ## Opens up - [Gaussian elimination and row echelon form](https://lightmysky.com/learn/mathematics/gaussian-elimination-and-row-echelon-form-mt_PZBce3UvRm)