C2.5; E2.5 - Changing the subject

C2.5; E2.5 - Changing the subject

A fare is calculated by C=4+3dC=4+3d. If the fare is known, how can you write a rule that gives the distance instead? C2.5 and E2.5 develop rearrangement by balanced inverse operations, including simple fractions; each requested subject appears once and has no power or root in this lesson.

Single operations

A formula describes a relationship between quantities. The opening C=4+3dC=4+3d already gives CC directly when dd is known. Changing its subject means expressing dd directly using CC, while keeping the same relationship.

Subject of a formula

The subject is the quantity isolated on one side of a formula, with coefficient 1. In P=4sP=4s, PP is the subject; writing s=P/4s=P/4 makes ss the subject instead.

Rearranging is equation solving with letters left in place. Apply the same operation to both sides and keep a complete equality on each line. Do not substitute numerical values unless the question asks for them: the aim is a rule valid for every allowed input.

Worked example 1: undoing addition

Question. Make rr the subject of p=r+sp=r+s.

Plan. Subtract ss from both sides to isolate rr.

p−s=r+s−sp−s=rr=p−swrite the isolated subject on the left\begin{aligned} p-s&=r+s-s\\ p-s&=r\\ r&=p-s &&\text{write the isolated subject on the left} \end{aligned}

Answer. r=p−sr=p-s.

Check. Replace rr in the original: (p−s)+s=p(p-s)+s=p. For instance, p=11p=11, s=4s=4 gives r=7r=7 and 7+4=117+4=11.

Change the original operation from addition to multiplication. Its inverse is division, which needs a non-zero divisor.

Worked example 2: undoing multiplication

Question. Make rr the subject of p=rsp=rs, where s≠0s\ne0.

Plan. Divide both sides by ss.

ps=rssps=rcancel the non-zero factor sr=ps\begin{aligned} \frac{p}{s}&=\frac{rs}{s}\\ \frac{p}{s}&=r &&\text{cancel the non-zero factor }s\\ r&=\frac{p}{s} \end{aligned}

Answer. r=psr=\frac{p}{s}, with s≠0s\ne0.

Check. Substituting gives (p/s)s=p(p/s)s=p. For p=12p=12, s=3s=3, the rearranged formula gives r=4r=4, and 4×3=124\times3=12.

The restriction matters: if s=0s=0, the original becomes p=0p=0, which either allows every rr or none. It cannot determine one value of rr by division.

Watch out

From p=r+sp=r+s, r=s−pr=s-p reverses the subtraction. From p=rsp=rs, r=p−sr=p-s undoes the wrong operation. Identify how the subject is joined to the other quantity before choosing the inverse.

Leave a formula as your answer.

Several operations

An operation chain is reversed from its last step. In y=5+3xy=5+3x, the input xx is multiplied by 3 and then 5 is added. Subtract 5 before dividing by 3. If you divide first, the division must apply to every term on each side.

You may work with the subject initially on either side of the equals sign. Equality has no preferred direction: 3x=y−53x=y-5 and y−5=3xy-5=3x state the same relationship.

Worked example 3: reversing two operations

Question. Make xx the subject of y=5+3xy=5+3x.

Plan. Undo the added 5, then undo multiplication by 3.

y−5=3xsubtract 5 from both sidesy−53=xdivide the whole side by 3x=y−53\begin{aligned} y-5&=3x &&\text{subtract 5 from both sides}\\ \frac{y-5}{3}&=x &&\text{divide the whole side by 3}\\ x&=\frac{y-5}{3} \end{aligned}

Answer. x=y−53x=\frac{y-5}{3}.

Check. 5+3(y−53)=5+y−5=y5+3\left(\frac{y-5}{3}\right)=5+y-5=y. The rearrangement reproduces the original output.

Change only the sign before 3x3x. The coefficient of xx is now −3-3, so the final division must include that sign.

Worked example 4: a negative coefficient

Question. Make xx the subject of y=5−3xy=5-3x.

Plan. Subtract 5 and divide by −3-3, then simplify the signs.

y−5=−3xx=y−5−3x=5−y3reverse both numerator signs\begin{aligned} y-5&=-3x\\ x&=\frac{y-5}{-3}\\ x&=\frac{5-y}{3} &&\text{reverse both numerator signs} \end{aligned}

Answer. x=5−y3x=\frac{5-y}{3}, or y−5−3\frac{y-5}{-3}.

Check. 5−3(5−y3)=5−(5−y)=y5-3\left(\frac{5-y}{3}\right)=5-(5-y)=y. At y=11y=11, the formula gives x=−2x=-2, and 5−3(−2)=115-3(-2)=11.

Continue this lesson.

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