Isolating a Variable

Isolating a variable means rearranging an algebraic equation so that a different variable is on its own. The goal is to choose a sequence of operations that will leave the variable of interest on one side and put all other terms on the other side of the equal sign. For example, making $x$ the subject of the formula $y = x + 5$ gives $x = y - 5.$ This is a critical technique for solving algebraic equations.

Algebraic equations can be solved by changing the subject of a formula, also called isolating variables.

Isolate $x$ in the following equation: $x + 4 = 12$.

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To isolate $x$, we must get rid of the $4$ term from the left side of the equation and move it to the right side of the equation. To do this, subtract $4$ from both sides of the equation: $x + 4 - 4 = 12 - 4$. Then we can see that $4 - 4 = 0$, so the left side of the equation is simply $x$. The resulting expression is $x = 8$. $_\square$

Sarah knows that Mark has two more bananas than he has apples. If Sarah knows that Mark has 4 bananas, how many apples does he have? Sarah figures out an algebraic expression to solve the problem: $A + 2 = B,$ where $A$ is the number of apples and $B$ is the number of bananas. How can Sarah use this equation to figure out how many apples Mark has?

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To figure out how many apples there are, we have to change Sarah’s equation into an equivalent equation of the form $A = \dots$. We can do this by subtracting $2$ from both sides of the equation since doing the same operation on both sides of the equation ensures that both sides are still equal: $$A + 2 - 2 = B - 2,\ \text{ which is equivalent to }\ A = B - 2.$$

Here we have isolated $A$. We know that Mark has $4$ bananas, so we can plug $4$ into our equation where $B$ appears: $A = 4 - 2 = 2.$ We see that Mark has 2 apples. This makes sense since Mark has $2$ more bananas than apples, and if Mark has $2$ apples, he must have $4$ bananas. $_\square$

The examples above combine several techniques for isolating variables in equations. Here is a more thorough explanation of these techniques.

Doing the same operation to both sides:
To ensure that both sides of the equation remain equal, if an operation is done to one side of the equation, it must be done to the other side of the equation as well. For example, to solve for $x$: $x + 2 = 3,$ the $+ 2$ on the left side of the equation must be removed. To do this, simply subtract $2$ from the left side of the equation. However, if $2$ is not also subtracted from the right side of the equation, a false statement arises. To make sure that the equation remains true, subtract $2$ from the right side of the equation. $x + 2 - 2 = 3 - 2 \implies x = 1$.

Inverse Operations:
To isolate the variable in question, cancel out (or undo) operations on the same side of the equation as the variable of interest while maintaining the equality of the equation. This can be done by performing inverse operations on the terms that need to be removed so that the variable of interest is isolated. Subtraction cancels out addition, and vice versa, and multiplication cancels out division, and vice versa.

The various methods are as follows:

$(1)$ Using subtraction to undo addition

Making $x$ the subject of $y=x+3$ requires isolating $x$ from all other terms. The subject is written on the left, so switch the sides to get $x$ on the left:

$$x+3=y. \qquad (\text{switch sides})$$

To isolate $x$, $x$ needs to be by itself on the left-hand side. But currently the left side of the equation is $x + 3.$ The inverse of addition is subtraction, so subtract 3 from the left side. In order to keep the equality true, it is necessary to subtract 3 from the right side as well, so subtract 3 from both sides:

$$\begin{aligned} x+3-3&=y-3 &(\text{subtract 3 from both sides}) \\ x&=y-3. &(\text{simplify}) \end{aligned}$$

$$$$ $(2)$ Using addition to undo subtraction

Making $x$ the subject of $y=x-5,$

$$\begin{aligned} x-5&=y &(\text{switch sides}) \\ x-5+5&=y+5 &(\text{add 5 to both sides}) \\ x&=y+5. &(\text{simplify}) \end{aligned}$$

$$$$ $(3)$ Using division to undo multiplication

Making $x$ the subject of $y=8x,$ we have

$$\begin{aligned} 8x&=y &(\text{switch sides}) \\ \frac{8x}{8}&=\frac{y}{8} &(\text{divide both sides by 8}) \\ x&=\frac{y}{8}. &(\text{simplify}) \end{aligned}$$

$$$$ $(4)$ Using multiplication to undo division

Making $x$ the subject of $y=\frac{x}{8},$ we have

$$\begin{aligned} \frac{x}{8}&=y &(\text{switch sides}) \\ \frac{x}{8}\cdot8&=8y &(\text{multiply both sides by 8}) \\ x&=8y. &(\text{simplify}) \end{aligned}$$

$$$$ $(5)$ Adding or subtracting a multiple of $x$ to collect terms

We have $$3x + 4x = 5,$$ which simplifies to $$7x = 5,$$ and to isolate $x$ do the following: $$x = \frac{5}{7}$$

For another example, the equation $$2x + 4 = x + 5$$ takes the following steps to isolate $x$:

$$\begin{aligned} 2x - x +4 &= x - x + 5\\ x + 4 &= 5\\ x + 4 - 4 &= 5 - 4\\ x &= 1. \end{aligned}$$

$$$$ $(6$) Taking the root of an exponent or exponentiating a root

Observe that $$x = y^2$$ simplifies to $$y = \pm \sqrt{x}.$$

Represent the equation $y = x + 7$ with subject $x.$

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We have

$$\begin{aligned} x+7&=y &(\text{switch sides}) \\ x+7-7&=y-7 &(\text{subtract 7 from both sides}) \\ x&=y-7. &(\text{simplify}) \end{aligned}$$

Thus, making $x$ the subject of the formula $y = x + 7$ gives $x = y - 7.$ $_\square$

Represent the equation $y=x-2$ with subject $x.$

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We have

$$\begin{aligned} x-2&=y &(\text{switch sides}) \\ x-2+2&=y+2 &(\text{add 2 to both sides}) \\ x&=y+2. &(\text{simplify}) \end{aligned}$$

Thus, making $x$ the subject of the formula $y = x - 2$ gives $x = y + 2.$ $_\square$

Represent the equation $y=2x$ with subject $x.$

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We have

$$\begin{aligned} 2x&=y &(\text{switch sides}) \\\\ \frac{2x}{2}&=\frac{y}{2} &(\text{divide both sides by 2}) \\\\ x&=\frac{y}{2}. &(\text{simplify}) \end{aligned}$$

Thus, making $x$ the subject of the formula $y = 2x$ gives $x = \frac{y}{2}.$ $_\square$

Represent the equation $y=\frac{x}{7}$ with subject $x.$

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We have

$$\begin{aligned} \frac{x}{7}&=y &(\text{switch sides}) \\\\ \frac{x}{7}\cdot7&=7y &(\text{multiply both sides by 7}) \\\\ x&=7y. &(\text{simplify}) \end{aligned}$$

Thus, making $x$ the subject of the formula $y = \frac{x}{7}$ gives $x = 7y.$ $_\square$

Some of these problems require skills in solving multi-step equations.

Setting Up Equations

Order of Operations

Multi-step Equations