Solving for a Variable

Intuition

A rectangle has area $15$ and length $5$ . What is its width? Divide the area by the length: $15 \div 5 = 3$ . That works whenever you know the area and the length, because $\text{width} = \text{area} \div \text{length}$ every time.

Now write that pattern as a formula instead of redoing the division each time. The area formula $A = lw$ gives the area once you already have the length and width. Rearranged once, it gives the width directly instead: $w = \dfrac{A}{l}$ . Check it against the rectangle above: $w = \dfrac{A}{l} = \dfrac{15}{5} = 3$ , the same width you already found by dividing. Rearranging a formula this way is the whole of this lesson. You do it with 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, 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 one fixed quantity, a stand-in for some length whose value you do not need to know while you work. 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#

The properties of equality do not care whether the multiplier you divide by is a numeral or a letter: they let you divide both sides by it 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 $w = \dfrac{A}{l}$ , provided $l \neq 0$ . 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 a fixed quantity is what makes a letter behave like the $4$ in $4w = 12$ .

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 the average rate over the whole trip. Solving instead for $t$ is the mirror image: dividing by $r$ gives $t = \dfrac{d}{r}$ , provided $r \neq 0$ , that is, the trip actually covers some distance.

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. Solving for $y$ takes no work at all, since $y$ already stands alone. 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, $y$ is treated as a fixed quantity, 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 $x$ , where $a \neq 0$ and $a$ , $b$ , $c$ are fixed 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. Try a numeric case first. In $3x + 2x = 20$ , both terms already hold $x$ , so combine them the way you combine like terms anywhere: $3x + 2x = 5x$ , which turns the equation into $5x = 20$ , so $x = 4$ . Writing $3x + 2x$ as $(3 + 2)x$ is exactly the distributive law read backward.

The same idea rescues a symbolic equation like $ax + bx = c$ , where the coefficients are letters instead of numbers. Dividing both sides by $a$ (assuming $a \neq 0$ ) does remove the coefficient from the first term, but the second term becomes $\dfrac{b}{a}x$ : $x$ is still sitting in two places, just with different coefficients now. No single division finishes the job until the two copies of $x$ are combined into one, the same combining step that turned $3x + 2x$ into $5x$ above. 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, provided $1 + rt \neq 0$ :

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

For an account earning interest, $r$ and $t$ are positive, so $1 + rt$ is always at least $1$ and the division is safe. 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 $x$ , where $m + n \neq 0$ .

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 a written number, you can see whether it is zero before you divide. With a letter, or an expression built from letters, you cannot see its value, so you name the condition instead: 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: a side length is positive, and for a trip that actually happens, its distance and duration are both positive, which forces its rate to be positive too. 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 $a \neq b$ . When the excluded value does happen, in this case $a = b$ , the rearranged formula simply does not apply, and the original equation needs its own look, separate from the rearranging done here.

Check your understanding

Solve $C = 2\pi r$ for $r$ . Which condition, if any, belongs with your answer?

r = \dfrac{C}{2\pi}

, and no condition is needed, because

2\pi

is a fixed number that is never zero B

r = \dfrac{C}{2\pi}

, provided

r \neq 0

r = \dfrac{C}{2\pi}

, provided

C \neq 0

r = \dfrac{2\pi}{C}

, provided

C \neq 0

Now try one where the divisor really is a letter. The volume of a box is $V = lwh$ , the length times the width times the height.

Check your understanding

Solve $V = lwh$ for $h$ . What condition must accompany your rearranged formula?

h = \dfrac{V}{lw}

, provided

lw \neq 0

h = \dfrac{V}{lw}

, provided

V \neq 0

h = \dfrac{lw}{V}

, provided

V \neq 0

h = \dfrac{V}{lw}

, and no condition is needed

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.

Now try it yourself. The first checkpoint rearranged $ax + b = c$ into $x = \dfrac{c - b}{a}$ . Pick $a = 2$ , $b = 1$ , $c = 7$ : compute $x$ from the rearranged formula, then substitute that value into the original $2x + 1 = 7$ and confirm both sides match.

Check your understanding

The formula $m = \dfrac{y - b}{x}$ , with $x \neq 0$ , 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 subtracting a term. Solving $y = mx + b$ for $x$ means subtracting $b$ from both sides, which gives $y - b$ , not $y + b$ . Apply the same operation, with the same sign, to both sides rather than sliding a term across unchanged.
Forgetting to factor when the target is in two terms. If $x$ appears in $ax + bx$ , dividing by $a$ (assuming $a \neq 0$ ) rescales both terms but does not isolate $x$ : it leaves $x + \dfrac{b}{a}x$ , with $x$ still in two places. 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)

Core practice