Solving for a Variable

Intuition

A formula is a machine pointed in one direction. The area formula $A = lw$ takes a length and a width and hands back an area. The formula is perfect until the day you know the area and one side and want the other. Rather than plug in numbers and re-solve from scratch every single time, you can rearrange the formula once, turning $A = lw$ into $w = \dfrac{A}{l}$ . That version now hands back a width on demand. That rearranging is the whole of this lesson. An equation with several letters in it is called a literal equation. You solve it for the letter you care about by the same balancing moves you already use on equations full of numbers. Whatever you do to one side you do to the other, until the letter you want stands alone. The one new wrinkle, and the interesting part, is what to do when the letter you are chasing shows up in two places at once.

Learning goals

Isolate a chosen letter, treating the others as fixed numbers
Rearrange $A = lw$ into $w = \dfrac{A}{l}$
Factor the target out when it appears in several terms
State the nonzero condition that dividing by a letter assumes
Check a rearranged formula with simple numbers from the original

What it means to solve for a variable

A literal equation is an equation that contains more than one letter. The word comes from the Latin littera, for letter, and once you start looking, literal equations are everywhere. The equations $A = lw$ , $d = rt$ , $P = 2l + 2w$ , and $C = 2\pi r$ are all literal equations, and so is nearly every formula in geometry and science.

To solve for a particular variable in such an equation is to rewrite the equation so that one variable stands alone on a side. Every other letter and number is gathered on the other side. Solving $A = lw$ for $w$ means reaching $w = \dfrac{A}{l}$ ; solving the same equation for $l$ means reaching $l = \dfrac{A}{w}$ . The relationship never changes what it says. You are only rewriting it so a different letter becomes the subject.

The key that unlocks all of this is a small shift in how you read the other letters. When you solve $A = lw$ for $w$ , treat $l$ as though it were a known number, a stand-in for some fixed length you simply have not been told. That is exactly the way the first lesson of this subject taught you to see a letter as a placeholder for a number. Then $A = lw$ has the same shape as $12 = 4w$ , and you already know to divide by the coefficient. Solving for a variable is nothing more than the equation solving of the last few lessons, carried out with letters sitting patiently where the numbers used to be.

The same moves, now with letters

Because the other letters behave like fixed numbers, the tools are the ones you already trust: the properties of equality. You may add, subtract, multiply, or divide both sides by the same quantity (nonzero for multiplying and dividing), and each move produces an equivalent equation. To solve for a target variable, undo the operations wrapped around it in reverse order, using inverse operations, until it stands alone. The only difference from before is that the numbers you add or divide by now wear letters.

Why isolating a letter uses the very same moves#

When you solved a numerical equation like $4w = 12$ , you divided both sides by $4$ to reach $w = 3$ . A literal equation is no different. In $A = lw$ the width $w$ is multiplied by the length $l$ , exactly as it was multiplied by $4$ before. The difference is that now the multiplier is a letter whose value you have not been told. The properties of equality do not care whether that multiplier is a numeral or a letter. They let you divide both sides by that multiplier all the same, provided it is not zero.

So divide both sides of $A = lw$ by $l$ . On the right, the $l$ in the numerator cancels the $l$ you divided by, leaving $w$ alone:

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

Read the result the usual way round and $w = \dfrac{A}{l}$ . Nothing here is new. The single move, dividing both sides by the coefficient of the target, is the same one that finished every equation in the last few lessons. Treating $l$ as though it were a fixed number is what makes a letter behave like the $4$ in $4w = 12$ . That is the whole idea behind solving for a variable.

The area formula solved for the width. A rectangle of length l and width w has area A, so A equals l times w. Dividing both sides by the length l isolates the width, giving w equals A over l.

Worked example 1 Solve $d = rt$ for the rate $r$

The distance formula $d = rt$ multiplies the rate $r$ by the time $t$ . To solve for $r$ , undo that multiplication by dividing both sides by $t$ , which a real trip guarantees is not zero:

\frac{d}{t} = \frac{rt}{t} = r, \qquad\text{so}\qquad r = \frac{d}{t}.

The rearranged formula reads rate as distance divided by time, which is exactly how a speedometer would compute it. Solving instead for $t$ is the mirror image: dividing by $r$ gives $t = \dfrac{d}{r}$ .

Rearranging a formula in more than one step

Most formulas wrap the target in more than one operation, so you peel them off in reverse order, just as you did solving two-step equations. Strip away anything added to the target first, then divide by whatever multiplies it. The whole routine from the equation-solving lesson carries over unchanged; the constants are simply letters now.

Worked example 2 Solve $P = 2l + 2w$ for the width $w$

The perimeter of a rectangle is $P = 2l + 2w$ . The target $w$ sits inside the term $2w$ , with $2l$ added on, so undo in reverse. Subtract the added $2l$ from both sides first:

P - 2l = 2w.

Now divide both sides by the coefficient $2$ . The division must reach the whole left side, every term of it:

w = \frac{P - 2l}{2}.

Watch that last step. Dividing only the $P$ and forgetting the $2l$ gives the wrong $\dfrac{P}{2} - 2l$ ; the correct result is $\dfrac{P - 2l}{2}$ , which equals $\dfrac{P}{2} - l$ once each term is divided.

Worked example 3 Solve $C = \dfrac{5}{9}(F - 32)$ for $F$

The formula $C = \dfrac{5}{9}(F - 32)$ turns a Fahrenheit temperature into Celsius. To aim it the other way and read Fahrenheit from Celsius, solve for $F$ . The target is under two layers: it is inside the parentheses, and the whole parenthesis is multiplied by $\frac{5}{9}$ . Undo the outer layer first by multiplying both sides by the reciprocal $\frac{9}{5}$ :

\frac{9}{5}C = F - 32.

Then add $32$ to both sides to free $F$ :

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

That is the companion formula every weather report uses. Rearranging once, rather than re-solving from scratch each time, is the whole payoff: one derivation hands you a formula aimed straight at Fahrenheit.

Worked example 4 Solve $y = mx + b$ for $x$

The equation $y = mx + b$ describes a line, and the graphing chapter will lean on it constantly. Solving for $y$ takes no work at all, since $y$ already stands alone, and that is exactly the form a grapher wants. Solving for $x$ takes the familiar two steps. Subtract $b$ from both sides:

y - b = mx.

Then divide by the coefficient $m$ , assuming the line is not horizontal so that $m \neq 0$ :

x = \frac{y - b}{m}.

Keep the whole numerator $y - b$ together over $m$ . Throughout, the value of $y$ is treated as a known number, no different from the constants $m$ and $b$ , and that is what lets the ordinary moves work.

Check your understanding

Solve $ax + b = c$ for $x$ , treating $a$ , $b$ , and $c$ as known constants.

x = \dfrac{c - b}{a}

x = \dfrac{c + b}{a}

x = \dfrac{c}{a} - b

x = \dfrac{b - c}{a}

When the target appears in more than one term

Every formula so far held the target in a single term. The one genuinely new situation is a target that appears in two or more terms at once, as in $ax + bx = c$ . You cannot divide the trouble away in one stroke, because $x$ is sitting in two places, and dividing by $a$ would clear it from the first term only. The fix is the factoring skill from the expanding lesson, the distributive law read in reverse.

Why a target in two terms must be factored out#

Suppose the variable you are solving for, $x$ , appears in two separate terms, as in $ax + bx = c$ . The distributive law says $x(a + b) = ax + bx$ . Read from right to left, that same equation says

ax + bx = x(a + b),

which gathers the two copies of $x$ into a single factor. This is the crucial move: once $x$ appears only once, multiplied by the quantity $a + b$ , one division finishes the job. Divide both sides by $a + b$ , which is legal as long as $a + b \neq 0$ :

x = \frac{c}{a + b}.

Factoring is not a trick saved for tidy expressions; it is the one step that turns two appearances of the target into one. And once the target appears a single time, the ordinary inverse operations isolate it exactly as before.

The same routine handles a target that starts on both sides of the equation. Bring every term containing the target to one side first, then factor it out.

Worked example 5 Solve $ax = bx + c$ for $x$

Here the target $x$ appears on both sides, in $ax$ and in $bx$ . First bring both $x$ terms to the same side by subtracting $bx$ from both sides:

ax - bx = c.

Now the two copies of $x$ sit together, so factor $x$ out by reading the distributive law backward:

x(a - b) = c.

With $x$ in a single place, divide both sides by $a - b$ , which requires $a \neq b$ so the divisor is not zero:

x = \frac{c}{a - b}.

Collecting the target on one side and factoring it out is the routine whenever the variable you want appears more than once.

Worked example 6 Solve $A = P + Prt$ for the principal $P$

The amount in a simple-interest account after a time $t$ is $A = P + Prt$ , the principal $P$ plus the interest $Prt$ it earns. To recover the principal from the final amount, solve for $P$ . The target appears in both terms on the right, so factor it out. Both $P$ and $Prt$ share the factor $P$ :

A = P(1 + rt).

Now $P$ multiplies the single quantity $1 + rt$ , so divide both sides by it:

P = \frac{A}{1 + rt}.

The factor pulled out was $1 + rt$ , not $rt$ . The lone $P$ is $P \cdot 1$ , so a $1$ stays behind inside the parentheses; dropping it and writing $P = \dfrac{A}{rt}$ is the most common slip here.

Check your understanding

Solve $mx + nx = p$ for $x$ .

x = \dfrac{p}{mn}

x = \dfrac{p}{m + n}

x = \dfrac{p - n}{m}

x = p - m - n

Dividing by a letter, and checking the result

Every division in this lesson, by $l$ , by $t$ , by $a + b$ , quietly assumed the divisor was not zero. With numbers that assumption is automatic. With letters it is a genuine condition, the same nonzero-denominator rule from the algebraic-fractions lesson. An expression you divide by, or place under a fraction bar, must not be zero, and any value that would make it zero is excluded. For the geometry and motion formulas above the condition takes care of itself, because a side length, a speed, and a duration are all positive. In a purely symbolic rearrangement you should name the condition out loud: $w = \dfrac{A}{l}$ holds provided $l \neq 0$ , and $x = \dfrac{c}{a - b}$ holds provided $a \neq b$ .

A rearranged formula is easy to check, and the check is worth the few seconds. Pick simple numbers that fit the original, then confirm the rearranged version agrees. Take $w = \dfrac{A}{l}$ . A rectangle with $l = 5$ and $w = 3$ has area $A = 15$ , and the rearranged formula returns $\dfrac{A}{l} = \dfrac{15}{5} = 3$ , matching the width you started with. When a test case comes out wrong, a sign or a stray term in the rearrangement is off. This test catches a slip exactly as substituting a numeric answer back into the original equation does.

Check your understanding

The formula $m = \dfrac{y - b}{x}$ can be rearranged. Solve it for $y$ .

y = mx + b

y = mx - b

y = \dfrac{x}{m} + b

y = m(x + b)

Common mistakes

Dividing only one term by the coefficient. After reaching $2w = P - 2l$ , dividing by $2$ must hit the entire other side: $w = \dfrac{P - 2l}{2}$ , which equals $\dfrac{P}{2} - l$ , never $\dfrac{P}{2} - 2l$ . Every term crosses the division, just as every term does when you distribute.
Losing a sign when moving a term. Solving $y = mx + b$ for $x$ means subtracting $b$ from both sides, so it arrives as $y - b$ , not $y + b$ . Track the sign of each term you move rather than sliding it across unchanged.
Forgetting to factor when the target is in two terms. If $x$ appears in $ax + bx$ , you cannot isolate it by dividing by $a$ ; that leaves $x$ still hiding in the second term. Collect the $x$ terms, factor $x$ out to $x(a + b)$ , and only then divide.
Pulling out the wrong factor. Factoring $P + Prt$ gives $P(1 + rt)$ , not $P(rt)$ : the lone $P$ is $P \cdot 1$ , so a $1$ stays inside. Check any factoring by expanding it back.
Dividing by a letter without noting it must be nonzero. The answer $x = \dfrac{c}{a - b}$ is valid only when $a \neq b$ . State the condition whenever you divide by an expression that contains a letter.
Solving for the wrong variable. “Solve for $w$ ” means $w$ ends up alone on one side with no other $w$ anywhere in the equation. Isolating $l$ instead, or leaving the target on both sides, does not answer the question that was asked.

Key takeaways

A literal equation has more than one letter, and most formulas are literal equations. To solve for a chosen variable is to isolate it on one side, treating every other letter as a fixed but unknown number.
The moves are the properties of equality from the last lessons, unchanged. Undo the operations wrapped around the target in reverse order, with inverse operations, so $A = lw$ becomes $w = \dfrac{A}{l}$ .
When the target appears in more than one term, collect those terms on one side and factor the target out, using the distributive law in reverse. Then divide by the quantity left behind: $ax + bx = c$ becomes $x(a + b) = c$ , so $x = \dfrac{c}{a + b}$ .
Dividing by a letter assumes it is nonzero, the same nonzero condition as a denominator; name the excluded values in a symbolic answer.
Check a rearranged formula by substituting simple numbers that fit the original.
Rearranging a formula lets you compute a different quantity directly, and solving a two-variable equation for $y$ is the setup for graphing a line in the next chapters.

Solving for a Variable

Learning goals

What it means to solve for a variable

The same moves, now with letters

Rearranging a formula in more than one step

When the target appears in more than one term

Dividing by a letter, and checking the result

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)