Unit 2, lesson 8

Rearranging Formulas

How to make any letter of a formula its subject using the same steps as for linear equations, while stating which divisors must be nonzero and handling the cases where they are zero.

  • Theory part 15
  • Practice part 16
  • Linear Equations

Theory, part 15

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Practice, part 16

A rectangle has area 24 square centimeters and length 6 centimeters. Cut it across its length into 6 strips, each 1 centimeter wide. The 24 square centimeters are shared equally, so each strip has area 4 square centimeters and is 4 centimeters long, which is the width. The formula A=lwA = lw was written to give the area, but here you needed the width. Rearranging a formula means turning it around so that it answers a different question. The SAT asks for this directly, often with answer choices that are all formulas.

See also Making a Variable the Subject in Foundations for Calculus.

Key terms

Formula

Definition A formula is an equation that states a general relationship among quantities, each named by a letter with a stated meaning and unit. A formula comes with the values its letters are allowed to take. For example, lengths and areas in a geometric figure are positive, and a time or a speed in a motion problem is usually nonnegative.

In plain words A formula is a recipe that holds for every case of a situation, not a single equation to be solved once.

Example For a rectangle with length ll and width ww in centimeters, the area in square centimeters is A=lwA = lw and the perimeter in centimeters is P=2l+2wP = 2l + 2w, where l>0l > 0 and w>0w > 0. A temperature of CC degrees Celsius is FF degrees Fahrenheit, where

F=95C+32.F = \frac{9}{5}C + 32.

Subject of a formula

Definition The subject of a formula is the letter that stands alone on one side of the equals sign and does not appear on the other side. Rearranging a formula, also called solving it for a letter, means rewriting it as an equivalent formula with that letter as the subject. Equivalent means that, for the allowed values of the letters, both formulas are true for exactly the same values.

In plain words The subject is the quantity the formula hands you directly. Changing the subject lets you compute a different quantity from the others.

Example In A=lwA = lw the subject is AA. With ww as the subject the formula reads

w=Al.w = \frac{A}{l}.

Isolate the letter first and substitute numbers afterwards. With A=24A = 24 and l=6l = 6 this gives w=4w = 4, and the same rearranged formula works for every other rectangle too.

Inverse operation

Definition To make a letter the subject, apply inverse operations to both sides of the formula, exactly as when solving a linear equation. Add or subtract the same expression on both sides, and multiply or divide both sides by the same nonzero expression. Undo the operations applied to the chosen letter in the reverse of the order in which they were applied, treating every other letter as a fixed number.

In plain words Pretend every other letter is a number you know. Then solve for your letter the way you would solve for xx.

Example Make CC the subject of F=95C+32F = \frac{9}{5}C + 32. The letter CC was multiplied by 95\frac{9}{5} and then had 3232 added, so subtract 3232 first and then multiply by 59\frac{5}{9},

F−32=95C,C=59(F−32).\begin{aligned} F - 32 &= \frac{9}{5}C, \\[0.4em] C &= \frac{5}{9}(F - 32). \end{aligned}

As a check, F=212F = 212 gives C=59⋅180=100C = \frac{5}{9} \cdot 180 = 100, the boiling point of water.

A sum works the same way. To make ww the subject of P=2l+2wP = 2l + 2w, subtract 2l2l and divide by 22,

w=P−2l2=P2−l.w = \frac{P - 2l}{2} = \frac{P}{2} - l.

Because w>0w > 0, this formula describes a real rectangle only when P>2lP > 2l.

These steps work because the chosen letter appears only to the first power. A formula such as A=s2A = s^2 for the area of a square needs a square root to make ss the subject, which is developed in Radicals and Rational Exponents once roots have been introduced.

Parameter

Definition A parameter is a letter that represents a fixed number within a particular problem, while other letters vary. In y=mx+by = mx + b the letters mm and bb are parameters, and each choice of them gives one line. When you rearrange a formula, parameters are treated like numbers, but their possible values still matter, because a parameter may turn out to be zero.

In plain words A parameter is a constant that has not been chosen yet. It stays put while xx and yy move, but you cannot assume it is a convenient number.

Example In the motion formula d=rtd = rt, a car on one trip has a fixed speed rr, so rr is a parameter while the time tt and the distance dd vary. At r=50r = 50 miles per hour the formula is d=50td = 50t, and after 3 hours the car has gone 150 miles.

Nonzero denominator

Definition Dividing both sides of an equation by an expression is allowed only for the values that make the expression nonzero, and the rearranged formula then has that expression as a denominator. State the restriction before you divide. In general, from ka=cka = c you may conclude

a=ckonly when k≠0.a = \frac{c}{k} \quad \text{only when } k \ne 0.

In plain words Every time you divide, ask what you are dividing by and whether it could be zero. Write the answer next to your formula.

Example Make tt the subject of the rate formula d=rtd = rt. Dividing by rr requires r≠0r \ne 0, so

t=dr,r≠0.t = \frac{d}{r}, \qquad r \ne 0.

For a moving car r>0r > 0, and the restriction holds automatically. A trip of 150 miles at 60 miles per hour takes t=15060=2.5t = \frac{150}{60} = 2.5 hours. Similarly w=Alw = \frac{A}{l} is safe for a rectangle, because a length is positive.

Exceptional parameter case

Definition When the coefficient of the chosen letter is a parameter or an expression that can be zero, the zero case must be handled separately. For the equation kx=ckx = c in xx, there are three possibilities.

CaseSolutions for xx
k≠0k \ne 0exactly one, x=ckx = \frac{c}{k}
k=0k = 0 and c=0c = 0every real number
k=0k = 0 and c≠0c \ne 0none

If the chosen letter appears in two terms, first collect those terms on one side and factor the letter out, so that it has a single coefficient. Then check whether that coefficient can be zero.

In plain words Dividing by a letter quietly assumes that letter is not zero. When it can be zero, look at that case on its own, where the formula either gives no answer or gives no information.

Example Solve ax+6=2x+bax + 6 = 2x + b for xx. Collect the xx terms and factor,

ax−2x=b−6(a−2)x=b−6.\begin{aligned} ax - 2x &= b - 6 \\[0.4em] (a - 2)x &= b - 6. \end{aligned}

If a≠2a \ne 2, then

x=b−6a−2.x = \frac{b - 6}{a - 2}.

If a=2a = 2, the equation reads 0=b−60 = b - 6. It has no solution when b≠6b \ne 6, and every real xx is a solution when b=6b = 6.

Factoring also helps in money formulas. With simple interest, a principal of PP dollars at an annual rate rr for tt years grows to A=P+PrtA = P + Prt dollars. Factoring out PP gives A=P(1+rt)A = P(1 + rt), and because r≥0r \ge 0 and t≥0t \ge 0, the factor 1+rt1 + rt is at least 11, so

P=A1+rtP = \frac{A}{1 + rt}

has no exceptional case. Solving for rr is different. It gives r=A−PPtr = \frac{A - P}{Pt}, which needs P≠0P \ne 0 and t≠0t \ne 0. At t=0t = 0 the formula says only A=PA = P, and no rate can be read from it.

Check yourself

  1. The formula v=v0+atv = v_0 + at gives the speed vv of an object after tt seconds, where v0v_0 is its starting speed and aa is a nonzero constant acceleration. Which equation gives tt in terms of vv, v0v_0 and aa? (A) t=v−v0at = \frac{v - v_0}{a} (B) t=va−v0t = \frac{v}{a} - v_0 (C) t=v+v0at = \frac{v + v_0}{a} (D) t=av−v0t = \frac{a}{v - v_0}
Show the answer

The correct choice is (A). Subtract v0v_0 to get v−v0=atv - v_0 = at, then divide by aa, which is allowed because a≠0a \ne 0. Choice (B) divides only vv by aa, choice (C) adds v0v_0 instead of subtracting it, and choice (D) turns the fraction upside down.

  1. The equation kx+3=5x+mkx + 3 = 5x + m, where kk and mm are constants and k≠5k \ne 5, is solved for xx. Which expression is equal to xx? (A) m+3k−5\frac{m + 3}{k - 5} (B) m−3k−5\frac{m - 3}{k - 5} (C) m−3k+5\frac{m - 3}{k + 5} (D) m−35−k\frac{m - 3}{5 - k}
Show the answer

The correct choice is (B). Collecting the xx terms gives kx−5x=m−3kx - 5x = m - 3, so (k−5)x=m−3(k - 5)x = m - 3. Since k≠5k \ne 5 the coefficient is not zero, and dividing gives the fraction in (B). Choice (A) adds 33 instead of subtracting it, (C) adds 55 to kk instead of subtracting it, and (D) has the right numerator but the opposite denominator, so it has the wrong sign whenever m≠3m \ne 3. If kk were 55, the equation would have no solution unless m=3m = 3.

  1. With simple interest, A=P(1+rt)A = P(1 + rt), where AA is the final amount in dollars, PP the principal in dollars, rr the annual rate as a decimal and tt the time in years. An account with P=1000P = 1000 grows to A=1150A = 1150 in t=3t = 3 years. What is the value of rr?
Show the answer

The answer is 0.050.05, which can be entered as 0.05, .05 or 1/20. Solving for rr with P≠0P \ne 0 and t≠0t \ne 0 gives

r=A−PPt=1503000=0.05.r = \frac{A - P}{Pt} = \frac{150}{3000} = 0.05.

The check gives 1000(1+0.15)=11501000(1 + 0.15) = 1150. Entering 55 would treat the rate as a percent, while the formula uses a decimal.

How this lesson connects