Solving for a Particular Variable

Need some basic practice with ‘undoing operations’ first? Undoing a Sequence of Operations

To ‘solve an equation for a particular variable’ means to rewrite the equation with the particular variable all by itself on one side.

The variable you are solving for cannot appear on the other side of the equation. Conventionally, the variable being solved for is put on the left-hand side of the equation.

Examples

Solving the equation $\quad PV = nRT\quad$ for $\quad P\quad$ gives:

$$P = \frac{nRT}{V}\quad$$

Solving the equation $\quad PV = nRT\quad$ for $\quad T\quad$ gives:

$$T = \frac{PV}{nR}\quad$$

The key to solving for a particular variable is to think about what is being done to the variable, and then to ‘undo’ these operations, in reverse order.

Question: Solve the equation $\quad ax + b = c\quad$ for $\quad x\,.$

Solution: Here is the correct thought process:

There is only one $\,x\,$ in the equation, in the expression $\,ax + b\,.$
The expression $\,ax + b\,$ represents a sequence of operations acting on $\,x\,$:
$$ \cssId{s22}{x} \quad\cssId{s23}{\overset{\text{multiply by $a$}}{\rightarrow}}\quad \cssId{s24}{ax} \quad\cssId{s25}{\overset{\text{add $b$}}{\rightarrow}}\quad \cssId{s26}{ax + b} $$
We must ‘undo’ these operations, in reverse order, to get back to $\,x\,.$ Note that addition is undone with subtraction, and multiplication is undone with division: $$ \cssId{s29}{ax + b} \quad\cssId{s30}{\overset{\text{subtract $b$}}{\rightarrow}}\quad \cssId{s31}{ax} \quad\cssId{s32}{\overset{\text{divide by $a$}}{\rightarrow}}\quad \cssId{s33}{x} $$

Thus, we start with the original equation, first subtract $\,b\,$ from both sides, and then divide both sides by $\,a\,$:

$ax + b = c$	(original equation)
$ax = c - b$	(subtract $b$ from both sides)
$\displaystyle x = \frac{c-b}{a}$	(divide both sides by $a$)

Solving for a Particular Variable

Examples

Concept Practice