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Rearranging Formulae When the Subject Appears Twice

Change the subject of a formula where the new subject shows up more than once: clear any fraction, collect those terms on one side, factorise the subject out, then divide.

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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 the bracket

1 · Read

The last stop showed you how to handle algebraic fractions: clear the fraction, factorise, cancel whole factors, and watch the excluded values. This stop uses those same moves to change the subject of a formula when the new subject appears more than once. The method has four moves: clear any fraction, collect every term containing the new subject on one side, factorise the subject out, and divide by the bracket.

Clear any fractionPut x terms on one sideFactorise x outDivide by the bracket
The four moves, in the order they happen.

Here is why the moves work. If x appears in two terms, you cannot just divide by x, because x is not a single factor yet. The trick is to make it one: collect every term containing x on one side and everything else on the other, then factorise x out. If x is inside a bracket, expand the bracket first. One more thing: the new denominator cannot be zero. In x = (2y + 1)/(y - 4), the value y = 4 would make the right side undefined, and the original formula has no solution there either.

Try it together

Make x the subject of y = (2x + 5)/(x - 3). Clear the fraction by multiplying both sides by (x - 3): y(x - 3) = 2x + 5. Expand: xy - 3y = 2x + 5. Collect the x terms on the left: xy - 2x = 3y + 5. Factorise x out: x(y - 2) = 3y + 5. Divide by the bracket: x = (3y + 5)/(y - 2). Check by substituting back: with x = (3y + 5)/(y - 2), the expression (2x + 5)/(x - 3) does simplify back to y.

y(x - 3) = 2x + 5xy - 3y = 2x + 5xy - 2x = 3y + 5x(y - 2) = 3y + 5x = (3y + 5)/(y - 2)
Try it together

Now x appears twice with no fraction. Make x the subject of A = 3x + 2xy. There is no fraction to clear, so go straight to collecting: every term already contains x, so nothing moves. Factorise x out: x(3 + 2y) = A. Divide by the bracket: x = A/(3 + 2y). The same four moves work here; two of them need nothing this time.

Try it together

Now x appears twice in denominators. Make x the subject of R = 1/x + 1/2. Clear the fractions by multiplying both sides by 2x: 2xR = 2 + x. Collect the x terms: 2xR - x = 2. Factorise x out: x(2R - 1) = 2. Divide: x = 2/(2R - 1). The value R = 1/2 is not allowed, because it makes the new denominator zero.

2xR = 2 + x2xR - x = 2x(2R - 1) = 2x = 2/(2R - 1)
Good to know

To check a rearrangement, substitute the finished expression back into the original formula and confirm it simplifies to the other side. It takes a few lines, but it catches a dropped term or a flipped sign before you use the formula.

Clear any fraction, collect the terms containing the new subject on one side, factorise it out, then divide by the bracket.

2 · Watch

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Rearranging Formulae When the Subject Appears Twice · Mathematics, ages 15 to 16 · LightMySky