10 min read

Rearranging a formula, and the errors that ruin calculus

Solving for whichever letter you want, and a tester that will tell you when an algebra step was a lie.

A weather feed gives temperatures in Fahrenheit and your program wants Celsius. The relationship you were given runs the wrong way: $F = \frac{9}{5}C + 32$ tells you $F$ from $C$ , and you need $C$ from $F$ . Guessing at the inverse is a good way to get an answer that is nearly right, so here is what nearly right looks like.

A rearrangement that is almost rightPython

No row agrees, and the gap between the two last columns is the same 14.2 degrees every time. What does a gap that never changes tell you about which operation the guess undid first? Then change the - 32 in the guess to - 17.7778 and run it again: every row is fixed, because $\frac{5}{9}\times 32 = 17.7778$ to four places.

The guess undid the multiplication before it undid the addition, and it is wrong by tens of degrees. Rearranging is not a matter of moving symbols about until it looks right; it is undoing operations in the correct order.

The method

A literal equation is one with several letters, where you are asked to solve for a chosen one. Treat the chosen letter as the only unknown and every other letter as though it were a number you happen not to know the value of. Then peel the operations off it in reverse order, doing the same thing to both sides each time.

Start from $F = \frac{9}{5}C+32$ , solving for $C$ .
The last thing done to $C$ was adding 32, so undo that first: subtract 32 from both sides, giving $F - 32 = \frac{9}{5}C$ .
The remaining operation on $C$ is multiplication by $\frac{9}{5}$ , so undo it: multiply both sides by $\frac{5}{9}$ .
That gives $C = \frac{5}{9}(F-32)$ . The bracket is compulsory, because the whole of $F-32$ was multiplied.
Check with a value you know: water boils at $C = 100$ and $F = 212$ . Then $\frac{5}{9}(212-32) = \frac{5}{9}\cdot180 = 100$ .

When the target appears in more than one place

Collect every term containing the target on one side, everything else on the other, then take the target out as a common factor and divide by what is left.

Solve $ax+b = cx+d$ for $x$ .

Move the $x$ terms together: subtract $cx$ from both sides, giving $ax - cx + b = d$ .
Move the rest away: subtract $b$ from both sides, giving $ax - cx = d - b$ .
The left side has $x$ in both terms, so factor it out: $x(a-c) = d-b$ .
Divide both sides by $(a-c)$ : $x = \dfrac{d-b}{a-c}$ .
This needs $a \ne c$ . If $a = c$ the original equation has no $x$ left in it at all, and it is either always true or never true.

A formula with the target underneath a fraction line needs the fractions cleared first, using the method from the lesson on fractions inside fractions. Solve $\dfrac{1}{c} = \dfrac{1}{a}+\dfrac{1}{b}$ for $a$ , a formula that describes resistors in parallel and lenses alike.

Get the term holding $a$ on its own: subtract $\frac{1}{b}$ from both sides, giving $\frac{1}{c}-\frac{1}{b} = \frac{1}{a}$ .
Combine the left side over the common denominator $bc$ : $\frac{b}{bc}-\frac{c}{bc} = \frac{b-c}{bc}$ .
So $\frac{b-c}{bc} = \frac{1}{a}$ .
Both sides are single fractions, so turn both upside down, which is legal as long as neither side is zero: $\frac{bc}{b-c} = a$ .
Check with numbers: $b = 6$ and $c = 2$ predict $a = \frac{12}{4} = 3$ . And indeed $\frac{1}{3}+\frac{1}{6} = \frac{1}{2}$ .

The errors, and a tester for them

An algebraic identity is a claim that two expressions agree at every value. That makes it easy to test and impossible to fudge: one value where they disagree kills the claim. This snippet is a tool worth keeping.

An identity testerPython

Change right to x ** 2 + 6 * x + 9 and run it again: zero disagreements. Then test a claim of your own, such as left returning (x + 3) ** 3 against a right you write yourself.

Every row of the table below is a claim that fails that test. They have names because they are so common, and each one is a way of pretending an operation spreads across a sum when it does not.

The false claim	A counterexample	What is actually true
$(a+b)^{2} = a^{2}+b^{2}$	$a=3, b=4$ : 49 against 25	$(a+b)^{2} = a^{2}+2ab+b^{2}$
$\sqrt{a+b} = \sqrt{a}+\sqrt{b}$	$a=9, b=16$ : 5 against 7	nothing simpler; leave it alone
$\dfrac{1}{a+b} = \dfrac{1}{a}+\dfrac{1}{b}$	$a=b=1$ : $\frac{1}{2}$ against 2	$\dfrac{1}{a}+\dfrac{1}{b} = \dfrac{a+b}{ab}$
$\dfrac{a+c}{b+c} = \dfrac{a}{b}$	$a=1,b=2,c=1$ : $\frac{2}{3}$ against $\frac{1}{2}$	only common factors cancel, never common terms
$\sqrt{a^{2}} = a$	$a=-5$ : 5 against $-5$	$\sqrt{a^{2}} = \|a\|$
$-(a-b) = -a-b$	$a=5,b=2$ : $-3$ against $-7$	$-(a-b) = -a+b$
$(ab)^{2} = ab^{2}$	$a=2,b=3$ : 36 against 18	$(ab)^{2} = a^{2}b^{2}$

Claims that are false, with one value that proves it.

Solve

A = P(1+rt)

for

t

Divide both sides by $P$ first, since $P$ multiplies the whole bracket: $\frac{A}{P} = 1+rt$ . Subtract 1: $\frac{A}{P}-1 = rt$ . Divide by $r$ : $t = \frac{\frac{A}{P}-1}{r}$ , which tidies by multiplying top and bottom by $P$ to $t = \frac{A-P}{Pr}$ . Check with $A = 120$ , $P = 100$ , $r = 0.1$ : the formula gives $t = \frac{20}{10} = 2$ , and $100(1+0.1\cdot2) = 120$ .

Check yourself

Solving $S = \dfrac{a}{1-r}$ for $r$ gives which of these?

r = 1 - \dfrac{a}{S}

r = \dfrac{a}{S} - 1

r = 1 - \dfrac{S}{a}

r = 1 - aS