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.
What a learner can do afterwards
- 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
1 · Read
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.
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.
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.
The original formula requires x != 3. The quotient requires y != 2.
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.
If the factor is zero, inspect the equation before division.
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.
The multiplier acts on both right-side terms.
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.
An original denominator must remain nonzero.
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.
The wanted letter determines which terms you collect.
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.
Changing both signs gives an equivalent quotient.
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.
These additional readings, practice questions and flashcards were written for this dataset.
More to read
A multiplier for every term
In an equation, a fraction-clearing multiplier applies to every term on both sides. Record the original denominator restrictions first.
Consider R = 2/x + 1/7, where x != 0. Multiplying by 7x gives 7xR = 14 + x. The product 7x(2/x) is 14, while 7x(1/7) is x.
Subtract x from both sides: 7xR - x = 14. Factorise the left side: x(7R - 1) = 14. When R != 1/7, divide to get x = 14/(7R - 1).
At R = 1/7, the equation before division becomes 0 = 14. It has no solution. For other values of R, the resulting x is nonzero, as the original denominator requires.
Check the zero factor separately
Factorising B = 5x + 2xt gives B = x(5 + 2t). When 5 + 2t is nonzero, the formula becomes x = B/(5 + 2t).
At t = -5/2, the factor multiplying x is zero. Return to B = 0x instead of dividing.
If B = 11, the equation says 11 = 0. No real x satisfies it. If B = 0, it says 0 = 0, so every real x satisfies it. The quotient is undefined in both cases, but the original equations have different solution sets.
Also avoid dividing by x without checking it. When B = 0 and t = 1, the formula is 0 = 7x, whose solution is x = 0. Dividing by x would discard that solution.
Make the other letter the subject
Start with x = (6y + 1)/(y - 4), where y != 4. To make y the subject, multiply by y - 4: xy - 4x = 6y + 1.
Subtract 6y and add 4x on both sides. This gives xy - 6y = 4x + 1. Factorise y: y(x - 6) = 4x + 1. For x != 6, divide to obtain y = (4x + 1)/(x - 6).
In the form y = (ax + b)/(x - c), the constants are a = 4, b = 1 and c = 6. The original restriction y != 4 still applies.
Check a pair of values. Taking y = 5 in the starting formula gives x = 31. The new formula returns (4(31) + 1)/(31 - 6) = 125/25 = 5.
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