6.02c - Changing the subject

6.02c - Changing the subject

A formula can answer a different question when you rearrange it. You will make a chosen letter the subject, including when it is squared, in a denominator or appears twice, and check which values the rearrangement permits.

Changing a subject that occurs once

A hire charge is given by C=4h+7C=4h+7. If you know the charge, how can you find the number of hours without guessing?

For a charge of £27, solve 27=4h+727=4h+7: subtract seven, then divide by four, giving five hours. Keeping CC as a letter gives a rule that works for every permitted charge. The balancing operations are exactly those used in a numerical equation.

Subject of a formula

The subject is the letter isolated on one side of a formula. In C=4h+7C=4h+7, CC is the subject. In h=(C−7)/4h=(C-7)/4, hh is the subject. Rearranging changes the form, while preserving the relationship for permitted values.

Undo the operations in reverse order. If a letter is multiplied and then increased, subtract before dividing. Apply each operation to both complete sides; letters represent numbers, so they obey the same rules.

Worked example 1: removing an added charge

Question. Make hh the subject of C=4h+7C=4h+7.

Plan. Undo adding seven, then undo multiplying by four.

C−7=4hsubtract 7C−74=hdivide by 4\begin{aligned} C-7&=4h &&\text{subtract 7}\\ \frac{C-7}{4}&=h &&\text{divide by 4} \end{aligned}

Answer. h=(C−7)/4h=(C-7)/4.

Check. Substitute the expression into the original: 4((C−7)/4)+7=C−7+7=C4((C-7)/4)+7=C-7+7=C. Also, C=27C=27 gives h=5h=5, as expected.

Change only the sign of the constant in the formula. The operation you undo changes with it.

Worked example 2: removing a subtraction

Question. Make hh the subject of C=4h−7C=4h-7.

Plan. Add seven before dividing by four.

C+7=4hadd 7h=C+74divide by 4\begin{aligned} C+7&=4h &&\text{add 7}\\ h&=\frac{C+7}{4} &&\text{divide by 4} \end{aligned}

Answer. h=(C+7)/4h=(C+7)/4.

Check. 4((C+7)/4)−7=C+7−7=C4((C+7)/4)-7=C+7-7=C. For C=13C=13, the new rule gives h=5h=5, and 4×5−7=134\times5-7=13.

Numerical coefficients can be replaced by letters. From y=ax+by=ax+b, subtraction gives y−b=axy-b=ax, so x=(y−b)/ax=(y-b)/a provided a≠0a\ne0. If a=0a=0, the original is y=by=b: every xx works when y=by=b, and none works otherwise. Dividing by zero would hide that distinction.

Watch out

(C−7)/4(C-7)/4 divides the whole difference by four. Writing C−7/4C-7/4 divides only the seven and does not undo 4h+74h+7.

Keep the whole numerator together in the next formula.

Subjects under powers and reciprocals

When the chosen letter is squared, first isolate its square. When it is in a denominator, multiply by that denominator before trying to isolate it.

For example, r2=9r^2=9 allows r=3r=3 and r=−3r=-3, because both square to nine. A radius, however, must be positive, so the context would select 33. The symbol 9\sqrt9 itself means the principal non-negative square root, 33; the two solutions come from writing ±9\pm\sqrt9.

Undoing a square

For real values, u2=ku^2=k gives u=±ku=\pm\sqrt{k} when k>0k>0, just u=0u=0 when k=0k=0, and no real solution when k<0k<0. Apply any contextual restrictions afterwards. This rule is not supplied on the 2027 formula sheets.

The next pair changes only the permitted sign of the subject; the algebraic relationship stays the same.

Worked example 3: both square roots

Question. Make tt the subject of s=3t2s=3t^2, where s≥0s\ge0 and tt may be any real number.

Plan. Isolate t2t^2, then retain both signs of its square root.

t2=s3divide by 3t=±s3undo the square\begin{aligned} t^2&=\frac{s}{3} &&\text{divide by 3}\\ t&=\pm\sqrt{\frac{s}{3}} &&\text{undo the square} \end{aligned}

Answer. t=±s/3t=\pm\sqrt{s/3}; at s=0s=0, both signs give the same value 00.

Check. Substitution gives 3(±s/3)2=s3(\pm\sqrt{s/3})^2=s. With s=12s=12, both t=2t=2 and t=−2t=-2 give twelve.

Now suppose tt is a time measured from the start. The condition t≥0t\ge0 selects one branch.

Worked example 4: selecting a permitted sign

Question. Make tt the subject of s=3t2s=3t^2, where s≥0s\ge0 and t≥0t\ge0.

Plan. Isolate the square as before, then use the stated non-negative condition.

t2=s3divide by 3t=s3use t≥0\begin{aligned} t^2&=\frac{s}{3} &&\text{divide by 3}\\ t&=\sqrt{\frac{s}{3}} &&\text{use }t\ge0 \end{aligned}

Answer. t=s/3t=\sqrt{s/3}.

Check. Its square is s/3s/3, and the chosen square root is non-negative. At s=12s=12, t=2t=2, whereas −2-2 is ruled out by the condition.

A reciprocal needs a different inverse operation. Multiplying by the subject's denominator removes that fraction bar; dividing by the other coefficient then isolates the subject.

Continue this lesson.

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