Rearrange nonlinear formulas

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Difficulty: Intermediate
Estimated time: 26 minutes
Domains: Advanced Math
Techniques: Formula-rearrangementSymbolic-isolationInverse-operationsSquare-root-branchesRestriction-preservation

What you’ll learn

Spot when a question wants a formula for one letter, not a number.
Treat every other letter as fixed while you get that one alone.
Undo the operations around it in a reliable order.
Pick the right sign after a square root, using a condition or the context.
Keep track of restrictions when the letter sits in a fraction or in more than one term.

Prerequisites

Why this matters on the SAT

Solve for the letter they ask for

A formula can have lots of letters, but the question asks about only one. That letter is your target. Think of every other letter as a number you already know: it stays in the answer, but you don’t solve for it. Your job is to peel away everything around the target until it stands alone.

Solution to the example

The target is $r$ . Right now $r^2$ is multiplied by $4\pi I$ , so divide both sides by $4\pi I$ :

r^2=\frac{P}{4\pi I}.

A square root undoes the square. Usually that gives two answers, one positive and one negative. But $r$ is a distance, and the problem says it’s positive, so keep the positive one:

r=\sqrt{\frac{P}{4\pi I}}.

The answer is C. Choice A stops one step early: it’s the formula for $r^2$ , not $r$ .

SAT example

The positive power $P$ received from a source is modeled by

P=4\pi Ir^2,

where $I$ is the positive intensity and $r$ is the positive distance from the source. Which equation expresses $r$ in terms of $P$ and $I$ ?

A
$r=\dfrac{P}{4\pi I}$
B
$r=\sqrt{\dfrac{4\pi I}{P}}$
C
$r=\sqrt{\dfrac{P}{4\pi I}}$
D
$r=\dfrac{\sqrt P}{4\pi I}$

Spot the words “in terms of”

The question usually tells you with words like express $y$ in terms of $x$ , solve the formula for $h$ , or which equation gives $v$ in terms of the other variables?

Those questions want a formula, not a number. So mark the letter they name, leave the others as they are, and aim for an answer with that letter by itself on one side.

Undo the outside first

Getting the target alone is like unwrapping a gift: the last layer on is the first layer off. So before you move anything, ask how the expression was built, starting at the target and working outward. Then undo those steps in reverse.

Say $b\ne0$ and

Q=a+b(c-x)^3,

and you want $x$ . Start at $x$ and build outward:

x \longrightarrow c-x \longrightarrow (c-x)^3 \longrightarrow b(c-x)^3 \longrightarrow a+b(c-x)^3.

Adding $a$ went on last, so it comes off first:

\begin{aligned} Q-a&=b(c-x)^3,\\[1.4em] \frac{Q-a}{b}&=(c-x)^3,\\[1.4em] \sqrt[3]{\frac{Q-a}{b}}&=c-x,\\[1.4em] x&=c-\sqrt[3]{\frac{Q-a}{b}}. \end{aligned}

Read down the lines: subtract $a$ , divide by $b$ (that’s why $b\ne0$ matters), take the cube root, then solve $c-x$ for $x$ . Each line does one thing to both whole sides. And $c-x$ stays grouped the whole time, so you never touch the inside while the outside is still wrapped around it.

Common mistake:

Taking the cube root too early. In $Q=a+b(c-x)^3$ , the cube covers only $c-x$ , not the whole right side, so a cube root right away doesn’t undo it. Subtract $a$ and divide by $b$ first. Then the cube stands alone, and the cube root undoes it.

Pick the valid branch of a square root

Square roots are where these questions get tricky. Say you reach

U^2=R.

For a real answer, $R$ can’t be negative, since no real number squares to a negative. When $R\ge0$ , the possibilities are

U=\pm\sqrt R.

Why the $\pm$ ? A number and its negative have the same square: $3^2$ and $(-3)^2$ are both $9$ . The symbol $\sqrt R$ on its own means only the nonnegative root, called the principal square root, so the $\pm$ brings the negative one back. Each sign gives one branch of the answer.

To choose a branch, look for a condition. If

(y-6)^2=R,

then

y=6\pm\sqrt R.

If $y\ge6$ , then $y-6$ is zero or positive, so take the plus branch: $y=6+\sqrt R$ .
If $y<6$ , then $y-6$ is negative, so take the minus branch: $y=6-\sqrt R$ .
If nothing points to one side, you may need both branches.

The context can choose too. A length, a speed, or an elapsed time is usually nonnegative. When that’s your reason, say it to yourself, rather than dropping the minus branch out of habit.

Common mistake:

Going from $(y-6)^2=R$ straight to $y-6=\sqrt R$ . That quietly assumes $y-6$ isn’t negative, and the square has hidden its sign. Write both branches first, $y-6=\pm\sqrt R$ , then let the condition or the context choose.

Example: Undo a square and select its branch

Worked example

The quantities $H$ , $c$ , $d$ , and $t$ satisfy

H=cd(t-2)^2,

where $c>0$ , $d>0$ , and $t<2$ . Which choice expresses $t$ in terms of $H$ , $c$ , and $d$ ?

A
$t=2+\sqrt{\dfrac{H}{cd}}$
B
$t=\sqrt{\dfrac{H}{cd}}-2$
C
$t=2-\dfrac{H}{cd}$
D
$t=2-\sqrt{\dfrac{H}{cd}}$

Step 1

Name the target

The question asks for $t$ , so that’s the target. $H$ , $c$ , and $d$ stay in the answer as fixed values.

Step 2

Get the square alone

The outside layer is the multiplication by $cd$ . Since $c$ and $d$ are both positive, $cd$ isn’t $0$ , so you can divide both sides by it:

\frac{H}{cd}=(t-2)^2.

Step 3

Write both branches

Take the square root of both sides. The condition comes next, so keep both signs for now:

t-2=\pm\sqrt{\frac{H}{cd}}.

Then add $2$ to both sides:

t=2\pm\sqrt{\frac{H}{cd}}.

Step 4

Let the condition choose

The problem says $t<2$ , so $t-2$ is negative. That’s the minus branch:

\boxed{t=2-\sqrt{\frac{H}{cd}}}.

The answer is D. Choice A is the plus branch, which would put $t$ at $2$ or above.

Step 5

Check by substituting

Want to be sure? Put your answer back into the original formula. From your answer,

t-2=-\sqrt{\frac{H}{cd}}.

cd(t-2)^2 =cd\left(-\sqrt{\frac{H}{cd}}\right)^2 =cd\left(\frac{H}{cd}\right) =H.

You get $H$ back, and $t$ sits on the side of $2$ the condition asks for. Notice that the minus sign disappears when you square. That’s exactly how the square hid it in the first place.

Factor the target when it appears more than once

Sometimes the target, or the same power of it, shows up in more than one term. Then there’s no single layer to peel off. Gather those terms on one side and factor first, so the target appears once.

Say $a+b\ne0$ and

E=ar^2+br^2.

Both terms have $r^2$ , so pull it out, then divide by $a+b$ :

\begin{aligned} E&=r^2(a+b),\\[1.4em] \frac{E}{a+b}&=r^2. \end{aligned}

If $r$ is a length, it’s nonnegative, so take the plus branch:

r=\sqrt{\frac{E}{a+b}}.

Dividing by only $a$ or only $b$ wouldn’t leave $r^2$ alone, because the other term would still be in the way. Factoring turns two target terms into one target factor.

Try it yourself:

From $M=pu^2-qu^2$ , write the nonnegative quantity $u$ in terms of $M$ , $p$ , and $q$ . Start by factoring $u^2$ out of the whole right side.

Check your understanding:

What formula did you get for $u$ , and what has to be true about the denominator?

Solve a fraction formula and keep its restrictions

Sometimes the target sits in both the top and the bottom of a fraction. The plan is three moves: clear the denominator, gather the target terms, and factor.

Say $a\ne0$ , $w+a\ne0$ , and

p=\frac{w-a}{w+a}.

The condition $w+a\ne0$ keeps the fraction defined. There’s a hidden restriction too: $p$ can’t be $1$ . If it were, the top and bottom would be equal, $w-a=w+a$ , and that only happens when $a=0$ . But $a\ne0$ .

Multiply both sides by the whole denominator:

\begin{aligned} p(w+a)&=w-a,\\[1.4em] pw+pa&=w-a. \end{aligned}

Now gather the $w$ -terms on one side and everything else on the other, then factor each side:

\begin{aligned} pa+a&=w-pw,\\[1.4em] a(p+1)&=w(1-p). \end{aligned}

The target appears once, so divide by $1-p$ :

\boxed{w=\frac{a(p+1)}{1-p}}.

The new denominator shows the restriction you found earlier, $p\ne1$ . It isn’t a side note: at $p=1$ this formula divides by $0$ , and the original formula can’t give $p=1$ anyway when $a\ne0$ .

Common mistake:

Trying to divide by $w$ before gathering the $w$ -terms. The target is in both $pw$ and $w$ , so dividing won’t get it alone. Move both terms to one side and factor first: $w-pw=w(1-p)$ . Then $w$ is a single factor, and dividing by $1-p$ leaves it by itself.

Check your understanding:

Which restrictions make $w=\frac{a(p+1)}{1-p}$ say exactly the same thing as the original formula, and why can’t $p$ be $1$ ?

Solve symbolic formulas by hand

When the answer is a formula in several letters, do the algebra by hand. Desmos can solve the equation once you give every other letter a number, but then you get one number, not the general formula. It also can’t tell you which branch or restriction the formula needs.

Plugging in sample values can still help. If you pick allowed values and your formula gives the wrong result, you’ve caught a mistake, either a copying slip or an algebra error. But a match doesn’t prove that two formulas agree for every allowed value, so treat it as a check, not a proof.

Practice problems

In each problem, find the target before you move anything.

Solve for a positive measurement

Practice problem

The area $A$ of a panel is modeled by

A=\frac32 bh^2,

where $b$ and $h$ are positive. Which equation expresses $h$ in terms of $A$ and $b$ ?

Calculator loads as you approach

Solve it by hand. If you like, try sample values here as a check.

Factor out a repeated cube

Practice problem

The real numbers $A$ , $m$ , $n$ , and $y$ satisfy

A=m(y-1)^3+n(y-1)^3,

where $m+n\ne0$ . Which equation expresses $y$ in terms of $A$ , $m$ , and $n$ ?

Calculator loads as you approach

Factor and rearrange by hand. Sample values here are an optional check.

Solve a mixture formula for the concentrate

Practice problem

In a coolant mixture, $c$ grams of concentrate is combined with $w$ grams of water. A concentration index $f$ is defined by

f=\frac{c}{2c+w},

where $c>0$ and $w>0$ . Which equation expresses $c$ in terms of $f$ and $w$ ?

Calculator loads as you approach

Clear the denominator and factor out c by hand. Sample values here are an optional check.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Read the last sentence first to find the target. Every other letter stays in the answer as a fixed value.
Undo the outside first, doing each step to both whole sides.
If the target or its power shows up in several terms, gather them and factor it out.
$U^2=R$ needs $R\ge0$ for a real answer. Write $U=\pm\sqrt R$ , then let the condition or the context pick the branch.
Keep denominator restrictions, and check a hard result by substituting it into the original formula.
Do these by hand. Sample values can catch a slip, but they can’t prove a general formula.

Related lessons

When an equation in one variable has the variable in a denominator and asks for numbers rather than a formula, see Solve rational equations.

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Solve radical equations

Use the same isolate-and-power idea to find possible solutions, then reject any value that fails the original radical equation.

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Practice

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151 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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