C2.5; E2.5 - Changing the subject

C2.5; E2.5 - Changing the subject

A formula can describe the same relationship with a different quantity on its own. You will reverse operations to change the subject, handle numerical and algebraic fractions, and check that a rearranged formula still represents the original relationship.

Reversing operations in a formula

A hire charge is C=5+3hC=5+3h, where hh is the number of hours. If you know the charge instead of the time, can one formula give the time straight away? Rearranging lets you ask a different question about the same relationship.

Subject of a formula

The subject is the letter isolated on one side of a formula, with no other occurrence of that letter on the other side. In C=5+3hC=5+3h, the subject is CC. To make hh the subject, write an equivalent formula with hh alone.

Solving an equation gave a particular number. Rearranging usually gives an expression in other letters. Treat these letters as quantities: subtract the same expression from each side, or divide both sides by the same non-zero quantity.

For C=5+3hC=5+3h, the operations on hh are “multiply by three, then add five”. Reverse their order: “subtract five, then divide by three”.

Worked example 1: reversing an added charge

Question. Make hh the subject of C=5+3hC=5+3h.

Plan. Remove the fixed charge first, then undo the multiplication.

C−5=3hsubtract fiveC−53=hdivide by threeh=C−53write the subject first\begin{aligned} C-5&=3h && \text{subtract five}\\ \frac{C-5}{3}&=h && \text{divide by three}\\ h&=\frac{C-5}{3} && \text{write the subject first} \end{aligned}

Answer. h=C−53h=\frac{C-5}{3}.

Check. If h=4h=4, the original gives C=17C=17. The new formula returns (17−5)/3=4(17-5)/3=4.

Change only the sign before 3h3h. This time the coefficient you divide by is negative.

Worked example 2: a negative coefficient

Question. Make hh the subject of C=5−3hC=5-3h.

Plan. Subtract five, then divide by negative three.

C−5=−3hsubtract fiveh=C−5−3divide by negative threeh=5−C3reverse numerator signs\begin{aligned} C-5&=-3h && \text{subtract five}\\ h&=\frac{C-5}{-3} && \text{divide by negative three}\\ h&=\frac{5-C}{3} && \text{reverse numerator signs} \end{aligned}

Answer. h=5−C3h=\frac{5-C}{3}.

Check. For h=1h=1, the original gives C=2C=2, and (5−2)/3=1(5-2)/3=1.

The second formula is an algebraic comparison; a hire charge normally rises with time, so the first formula models that context. Always preserve what a supplied formula actually says rather than changing its signs to match a guess.

Watch out

C=5+3hC=5+3h does not give h=C−53h=C-\frac53. The entire difference C−5C-5 is divided by three: h=C−53h=\frac{C-5}{3}. The fraction bar acts like brackets.

Writing the subject on the left is conventional; equality itself works in either direction.

Changing a subject through fractions

A fraction bar groups its numerator. In p=q+rsp=\frac{q+r}{s}, the whole sum q+rq+r is divided by ss. Multiplying both sides by ss clears that division. The original formula requires s≠0s\ne0.

When the letter to be isolated is a denominator, first multiply to bring it out of the denominator. Later division by another letter needs its own non-zero condition. These formulas still have the target appearing only once, without a power or root.

Worked example 3: the subject is in a numerator

Question. Make qq the subject of p=q+rsp=\frac{q+r}{s}, where s≠0s\ne0.

Plan. Multiply by ss before subtracting rr.

ps=q+rmultiply by sps−r=qsubtract rq=ps−rwrite q first\begin{aligned} ps&=q+r && \text{multiply by }s\\ ps-r&=q && \text{subtract }r\\ q&=ps-r && \text{write }q\text{ first} \end{aligned}

Answer. q=ps−rq=ps-r, with the original condition s≠0s\ne0.

Check. Take p=5, r=2, s=3p=5,\ r=2,\ s=3. The new formula gives q=13q=13, and (13+2)/3=5(13+2)/3=5 in the original.

Keep the formula unchanged and ask for a different subject: ss. The target now starts below the fraction bar.

Worked example 4: the subject is a denominator

Question. Make ss the subject of p=q+rsp=\frac{q+r}{s}, for p≠0p\ne0 and s≠0s\ne0.

Plan. Clear the fraction, then divide by the non-zero coefficient pp.

ps=q+rmultiply by ss=q+rpdivide by p\begin{aligned} ps&=q+r && \text{multiply by }s\\ s&=\frac{q+r}{p} && \text{divide by }p \end{aligned}

Answer. s=q+rps=\frac{q+r}{p}. Retain p≠0p\ne0 and the original restriction s≠0s\ne0.

Check. With p=5, q=13, r=2p=5,\ q=13,\ r=2, the formula gives s=3s=3; the original gives 5=(13+2)/35=(13+2)/3.

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