Equations and Rearranging Formulas

Treat an equation as a statement that two expressions are equal. Preserve that equality while isolating a quantity, state restrictions on each operation, and check a result in the original equation.

An equation states that two expressions have the same value. Solving an equation means finding every allowed value that makes the statement true. Rearranging a formula uses the same algebraic principles to express one quantity in terms of the others.

Preserve Equality

Consider the equation:

2x+5=17

The value $x=6$ makes the equation true, while $x=5$ does not. The set of all values that make an equation true is its solution set.

We can preserve the solution set by applying a reversible operation to both sides. Subtract $5$ from both sides:

2x+5-5=17-5,

which simplifies to:

2x=12.

Then divide both sides by the nonzero number $2$ :

x=\frac{12}{2}=6.

These moves undo the operations applied to $x$ in reverse order. Substituting $x=6$ into the original equation confirms that $2(6)+5=17$ .

Exercise: Solve a linear equation

Solve $4x-3=21$ .

Your answer

Compute it first, then check your number.

HintUndo operations in reverse

First undo subtraction by adding $3$ . Then undo multiplication by dividing by $4$ .

SolutionKeep both sides equal

Add $3$ to both sides:

4x=24

Divide both sides by $4$ :

x=6

Substitution checks the answer: $4(6)-3=21$ .

Rearranging a Formula

A formula may contain several named quantities rather than one unknown number. To isolate a quantity is to write it alone on one side of the equality. Suppose:

y=wx+b

To isolate $x$ , subtract $b$ from both sides:

y-b=wx

and then divide both sides by $w$ :

x=\frac{y-b}{w}

The division step requires $w\ne0$ . If $w=0$ , the original formula becomes $y=b$ and no longer determines a unique value of $x$ . Stating this restriction is part of a complete rearrangement.

Exercise: Recover an input from a linear rule

For $y=3x+2$ , what value of $x$ produces $y=14$ ?

Your answer

Compute it first, then check your number.

HintUse the rearranged formula

Isolate the input: $x=(y-2)/3$ .

SolutionUndo the bias and scale

Substitute $14$ for $y$ :

x=\frac{14-2}{3}=\frac{12}{3}=4

Substituting $x=4$ into the original rule gives $3(4)+2=14$ , so the recovered input is correct.

Equivalent Forms Reveal Different Properties

The expressions

(x-2)(x+2)

and

x^2-4

have the same value for every $x$ . We can verify this by using the distributive law:

\begin{aligned} (x-2)(x+2) &=x(x+2)-2(x+2)\\ &=x^2+2x-2x-4\\ &=x^2-4. \end{aligned}

An equality that is true for every allowed value of its variables is an identity. By contrast, $x^2-4=0$ is true only for particular values of $x$ , so it is an equation to solve.

Different equivalent forms reveal different properties. The factored form $(x-2)(x+2)$ makes the zeros $x=2$ and $x=-2$ easy to see. The expanded form $x^2-4$ makes the highest power and constant term easy to see. Expanding removes products of grouped expressions; factoring reverses that process.

Algebra is therefore not only a way to obtain an answer. It is a way to choose a form that exposes the property we need.

Exercise: Recognize an identity

Which equality is true for every real value of $x$ ?

Choose the identity

2x + 1 = 7

3(x + 2) = 3x + 6

x² = 4

Select one choice, then check.

HintTest the meaning of every

An equation with one or two solutions is not an identity. Look for an algebraic rule that preserves equality for every $x$ .

SolutionApply the distributive law

Distributing $3$ gives:

3(x+2)=3x+6.

This equality holds for every real $x$ . The equation $2x+1=7$ has only the solution $x=3$ , and $x^2=4$ has only the solutions $x=2$ and $x=-2$ .

Keep Track of Restrictions

Not every operation is reversible for every value. A transformation can lose a solution or introduce a value that did not solve the original equation:

dividing by an expression can lose cases in which that expression is zero;
squaring both sides can introduce an extra candidate solution;
taking a square root requires attention to both possible signs when solving a squared equation;
applying a logarithm requires a positive input.

For example, $x^2=9$ has two real solutions:

x=3 \quad\text{or}\quad x=-3

The symbol $\sqrt{9}$ denotes the principal, nonnegative square root $3$ , but solving $x^2=9$ asks for every number whose square is $9$ . Those are two different questions.

Exercise: Check both square solutions

Which values solve $(x-1)^2=16$ ?

Choose the complete solution

Only x = 5

Only x = -3

x = 5 and x = -3

Select one choice, then check.

HintUndo the square in both directions

If $u^2=16$ , then $u=4$ or $u=-4$ . Here $u=x-1$ .

SolutionKeep both possible signs

From $(x-1)^2=16$ :

x-1=4 \quad\text{or}\quad x-1=-4

Therefore:

x=5 \quad\text{or}\quad x=-3

Substituting either value makes the squared expression equal $16$ .

Exercise: Find a solution lost by division

The equation

x(x-2)=0

has the solutions $x=0$ and $x=2$ . If we divide both sides by $x$ , which solution is lost?

Your answer

Compute it first, then check your number.

HintInspect the divisor

Division by $x$ is valid only when $x\ne0$ .

SolutionSeparate the zero case

Dividing by $x$ changes the equation to $x-2=0$ , which gives only $x=2$ . The division assumed $x\ne0$ , so it removed the valid solution $x=0$ .

Factoring avoids this loss: a product is zero when at least one factor is zero, so $x=0$ or $x-2=0$ .

Transform, State, and Check

A reliable algebraic method has three parts. Apply a valid operation consistently, state any restriction introduced by that operation, and substitute each candidate solution into the original equation. The final check is especially important when an earlier step was not reversible for every possible value.