Rearranging Formulae

To make \( x \) the subject, we must undo the operations done to \( x \) in reverse order. In \( y = 5x \; – \; 3 \), the operations are: multiply by 5, then subtract 3. Reversing: add 3 first, giving \( y + 3 = 5x \), then divide by 5, giving \( x = \frac{y + 3}{5} \).

We can verify: if \( x = 2 \), then \( y = 5(2) \; – \; 3 = 7 \), and \( \frac{7 + 3}{5} = \frac{10}{5} = 2 \). ✓

A common error is to divide by 5 first, getting \( \frac{y}{5} = x \; – \; 3 \), then add 3 to get \( x = \frac{y}{5} + 3 \). This is actually a valid approach when done correctly — but students who try it often forget to divide the −3 by 5 as well. The key “wrong order of inverse operations” misconception is thinking you can undo operations in any order without adjusting the other terms.

In \( s = vt \), the variable \( v \) is being multiplied by \( t \). The inverse of multiplication is division, not subtraction. Dividing both sides by \( t \): \( v = \frac{s}{t} \). We can verify: if \( v = 6 \) and \( t = 3 \), then \( s = 18 \). Checking: \( \frac{s}{t} = \frac{18}{3} = 6 = v \) ✓, but \( s \; – \; t = 18 \; – \; 3 = 15 \neq 6 \) ✗.

The “inverse of multiplication is subtraction” misconception arises because students sometimes see “getting rid of” a term as always meaning subtraction. When two variables are written next to each other (like \( vt \)), students may not recognise this as multiplication and so apply the wrong inverse. The key principle: if a variable is multiplied, you divide to undo; if it is added, you subtract to undo.

Starting from \( y = \frac{x}{5} + 2 \), subtract 2 from both sides: \( y \; – \; 2 = \frac{x}{5} \). Now multiply both sides by 5. The crucial point is that we must multiply the entire left-hand side by 5, not just the \( y \): \( 5(y \; – \; 2) = x \), so \( x = 5(y \; – \; 2) = 5y \; – \; 10 \). The wrong answer \( x = 5y \; – \; 2 \) comes from only multiplying the \( y \) by 5 and leaving the −2 unchanged.

This is the “partial application of an operation” misconception. When multiplying both sides by 5, students must multiply every term on each side. A quick check confirms: if \( x = 20 \), then \( y = \frac{20}{5} + 2 = 6 \). The correct formula gives \( 5(6 \; – \; 2) = 5 \times 4 = 20 \) ✓, while the wrong formula gives \( 5(6) \; – \; 2 = 28 \) ✗.

The expression \( 3x + 1 \) is all divided by 2, so we start by multiplying both sides by 2: \( 2y = 3x + 1 \). Next, subtract 1: \( 2y \; – \; 1 = 3x \). Finally, divide by 3: \( x = \frac{2y \; – \; 1}{3} \). We can verify: if \( x = 3 \), then \( y = \frac{3(3) + 1}{2} = \frac{10}{2} = 5 \), and \( \frac{2(5) \; – \; 1}{3} = \frac{9}{3} = 3 \). ✓

A common error is to multiply only part of the fraction by 2, or to treat the numerator’s terms separately — for instance, thinking “multiply \( x \) by 2 and leave the 1 alone.” The correct first step recognises the fraction bar as a grouping symbol: everything above the line is divided by 2, so everything gets multiplied when we clear the denominator. This is the “incorrect handling of nested operations” misconception.

The wrong answer leaves the subject \( x \) on both sides of the equals sign. To make \( x \) the subject, it must only appear once in the final expression. We must collect all \( x \) terms on one side: subtract \( cx \) and subtract 3 from both sides to get \( ax \; – \; cx = 5 \).

The crucial “gateway” step is to factorise out the \( x \) to combine the terms: \( x(a \; – \; c) = 5 \). Finally, divide both sides by the bracket: \( x = \frac{5}{a \; – \; c} \). The incorrect answer falls into the “subject on both sides” trap, which fails the fundamental definition of changing the subject of a formula.

Example: \( y = 3x + 7 \) — subtract 7: \( y \; – \; 7 = 3x \), then divide by 3: \( x = \frac{y \; – \; 7}{3} \)

Another: \( P = 5t + 2 \) — subtract 2: \( P \; – \; 2 = 5t \), then divide by 5: \( t = \frac{P \; – \; 2}{5} \)

Creative: \( c = 100a + 50 \) — subtract 50: \( c \; – \; 50 = 100a \), then divide by 100: \( a = \frac{c \; – \; 50}{100} \). Uses large numbers but the same structure.

Trap: \( y = 3(x + 7) \). A student might think “there’s a +7 so I must subtract first,” but here the 3 multiplies the entire bracket. You can divide by 3 first: \( \frac{y}{3} = x + 7 \), then subtract 7: \( x = \frac{y}{3} \; – \; 7 \). The “rigid procedural order” misconception assumes you must always subtract before dividing, ignoring the structure of the formula.

Example: \( y = x^2 \) — take the square root: \( x = \pm\sqrt{y} \)

Another: \( A = \pi x^2 \) — divide by \( \pi \), then root: \( x = \sqrt{\frac{A}{\pi}} \)

Creative: \( y = 3x^2 + 1 \) — subtract 1, divide by 3, then root: \( x = \pm\sqrt{\frac{y \; – \; 1}{3}} \). This requires three steps before the square root appears.

Trap: \( y = \sqrt{x} \). A student might think “there’s a square root in the formula, so making \( x \) the subject must also involve a square root.” But the inverse of a square root is squaring, not another square root. The correct rearrangement is \( x = y^2 \). This is the “getting the inverse the wrong way around” misconception.

Example: \( y = \frac{k}{x} \) — multiply both sides by \( x \): \( xy = k \), then divide by \( y \): \( x = \frac{k}{y} \). Now \( y \) is in the denominator.

Another: \( v = \frac{D}{t} \) — making \( t \) the subject: \( t = \frac{D}{v} \). The variable \( v \) ends up in the denominator.

Creative: \( m = \frac{F}{a} \) — making \( a \) the subject: \( a = \frac{F}{m} \). Here \( m \) appears in the denominator, even though \( a \) was originally in the denominator.

Trap: \( y = \frac{x}{3} \) — multiply by 3: \( x = 3y \). A student might think “there’s already a fraction, so the rearranged version must also have a variable in the denominator.” But the denominator is just the constant 3, and rearranging eliminates the fraction entirely. The condition requires a variable (not a constant) in the denominator.

Example: \( y = ax + bx \) — factorise: \( y = x(a + b) \), then divide: \( x = \frac{y}{a + b} \)

Another: \( ax + 3 = cx + d \) — collect \( x \) terms: \( ax \; – \; cx = d \; – \; 3 \), factorise: \( x(a \; – \; c) = d \; – \; 3 \), then \( x = \frac{d \; – \; 3}{a \; – \; c} \)

Creative: \( y(x + 1) = 3x \) — expand: \( yx + y = 3x \), collect: \( yx \; – \; 3x = -y \), factorise: \( x(y \; – \; 3) = -y \), so \( x = \frac{y}{3 \; – \; y} \). This requires expanding brackets first to reveal the two \( x \) terms.

Trap: \( y = x^2 + 3x \). A student might factorise as \( y = x(x + 3) \) and then write \( x = \frac{y}{x + 3} \), but this fails because \( x \) still appears on the right-hand side. The formula is actually a quadratic in \( x \) and requires the quadratic formula to solve. The “factorising always isolates the subject” misconception breaks down when different powers of the variable are involved.

This is the fundamental principle of rearranging. An equation is a balanced statement — whatever operation you perform on one side, you must perform on the other to maintain the equality. If \( y = 3x + 5 \) and you subtract 5 from the left but not the right, you get \( y \; – \; 5 = 3x + 5 \), which is false.

Students who “move” terms from one side to the other without thinking about balance are using a shortcut that disguises this principle. The shortcut works when applied correctly, but the “not applying to both sides” error leads to fundamentally wrong rearrangements.

If \( x \) is positive, then \( x = \sqrt{y} \) is correct — for example, \( x = 3 \), \( y = 9 \), \( \sqrt{9} = 3 \) ✓. But if \( x \) is negative, this fails — for example, \( x = -3 \), \( y = 9 \), \( \sqrt{9} = 3 \neq -3 \) ✗. The complete rearrangement is \( x = \pm\sqrt{y} \).

This targets the classic “forgetting the ±” misconception. The square root symbol (√) denotes only the positive root by convention. When rearranging, we must include both possibilities because squaring a negative number also gives a positive result. Students who write \( x = \sqrt{y} \) without the ± are losing solutions.

The correct rearrangement is: \( y = a \; – \; x \) → add \( x \) to both sides: \( y + x = a \) → subtract \( y \): \( x = a \; – \; y \). The statement claims \( x = y \; – \; a \). These are equal only when \( a \; – \; y = y \; – \; a \), which gives \( 2a = 2y \), so \( y = a \) (and then \( x = 0 \)).

TRUE case: \( a = 5 \), \( y = 5 \). Correct: \( x = 5 \; – \; 5 = 0 \). Claimed: \( x = 5 \; – \; 5 = 0 \). Same ✓. FALSE case: \( a = 5 \), \( y = 2 \). Correct: \( x = 5 \; – \; 2 = 3 \). Claimed: \( x = 2 \; – \; 5 = -3 \). Different ✗. This targets the “subtraction is commutative” misconception — students who think \( a \; – \; y \) and \( y \; – \; a \) are interchangeable. Unlike addition, subtraction is not commutative: the order matters.

The inverse of squaring is taking the square root, not halving. Halving is the inverse of doubling. Students who confuse “squared” with “times 2” will try to halve instead of square-root when rearranging formulae with squared terms. For example, from \( y = x^2 \), the correct rearrangement is \( x = \pm\sqrt{y} \), not \( x = \frac{y}{2} \).

This is the “squaring means doubling” misconception. It may arise because \( 2^2 = 4 \) and \( 2 \times 2 = 4 \), making squaring and doubling look identical. But \( 5^2 = 25 \) while \( 5 \times 2 = 10 \) — clearly different. Similarly, \( 10^2 = 100 \) but \( 10 \times 2 = 20 \). The operations only give the same result for 0 and 2, which is a coincidence rather than a rule.

The student’s first step is correct: dividing by 3 gives \( \frac{y}{3} = x + 2 \). The error is in the second step. The +2 is being added to \( x \), so to undo it we must subtract 2 from both sides: \( x = \frac{y}{3} \; – \; 2 \). The student kept the +2 as positive when it should have become −2.

This is the “sign error when rearranging” misconception — forgetting to change the sign of a term when it “moves” across the equals sign. A quick check: if \( x = 1 \), then \( y = 3(1 + 2) = 9 \). Correct: \( \frac{9}{3} \; – \; 2 = 1 \) ✓. Student’s answer: \( \frac{9}{3} + 2 = 5 \) ✗.

The answer is correct, but the reasoning reveals the “moving terms” misconception. The student thinks of rearranging as physically moving pieces around the equation rather than applying inverse operations to both sides. They say the 2 “goes underneath” when it crosses the equals sign, rather than understanding they are dividing both sides by 2.

For this simple case, the shortcut produces the right answer. But “moving” breaks down with more complex formulae — for instance, with \( y = (2x + 1)^2 \), a student who “moves” terms would struggle to know which operations to undo and in what order. The correct reasoning is: add \( 2x \) to both sides → \( y + 2x = 10 \); subtract \( y \) → \( 2x = 10 \; – \; y \); divide both sides by 2 → \( x = \frac{10 \; – \; y}{2} \).

The student’s error is in the first step. Subtracting 1 from \( y \) does not subtract 1 from inside the square root. The expression \( \sqrt{2x + 1} \; – \; 1 \) is not the same as \( \sqrt{2x + 1 \; – \; 1} = \sqrt{2x} \). The square root applies to the entire expression \( 2x + 1 \) as a single unit — you cannot “reach inside” and remove terms.

Let’s test the student’s logic with numbers we know. We know \( \sqrt{25} = 5 \). The student’s logic says that \( \sqrt{25} \; – \; 9 \) is the same as \( \sqrt{25 \; – \; 9} \). But \( \sqrt{25} \; – \; 9 = 5 \; – \; 9 = -4 \). Meanwhile, \( \sqrt{25 \; – \; 9} = \sqrt{16} = 4 \). Since \( -4 \neq 4 \), we prove we cannot pull operations inside a square root!

This is the “operating inside a function” misconception. The correct approach: square both sides first to remove the root: \( y^2 = 2x + 1 \). Then subtract 1: \( y^2 \; – \; 1 = 2x \). Then divide by 2: \( x = \frac{y^2 \; – \; 1}{2} \).

The student has confused multiplication with addition. In the formula \( A = \pi r^2 \), the \( \pi \) is multiplying \( r^2 \), not being added to it. The inverse of multiplication is division, not subtraction. The correct first step is to divide both sides by \( \pi \): \( \frac{A}{\pi} = r^2 \). Then take the square root: \( r = \sqrt{\frac{A}{\pi}} \).

This is the “inverse of multiplication is subtraction” misconception — the student sees “π next to \( r^2 \)” and treats it as an additive term rather than a multiplicative one. Check with \( r = 3 \): \( A = 9\pi \approx 28.27 \). Correct: \( \sqrt{\frac{28.27}{\pi}} = \sqrt{9} = 3 \) ✓. Student’s: \( \sqrt{28.27 \; – \; \pi} \approx \sqrt{25.13} \approx 5.01 \) ✗.

The student correctly subtracted 12, but they forgot that the minus sign belongs to the \( x \). The equation becomes \( y \; – \; 12 = -x \), not \( x \). The student simply dropped the negative sign to make their life easier.

To fix this, they can either multiply everything by −1 to get \( -y + 12 = x \) (which rearranges to \( x = 12 \; – \; y \)), or, better yet, they could avoid the negative altogether by adding \( x \) to both sides first: \( y + x = 12 \), then subtracting \( y \): \( x = 12 \; – \; y \). This targets the highly common “dropping the negative” misconception.