Solving One-Step Equations

Intuition

You already know that an equation like $x + 5 = 12$ is a sentence claiming two expressions are equal. You also know that a solution is a number that makes the sentence true. Here the solution is $7$ , because $7 + 5 = 12$ . But staring at the equation until the answer occurs to you is not a method, and it stops working the moment the numbers get awkward. This lesson gives you the actual method. Picture the equation as a balanced scale: the left side and the right side weigh exactly the same. If you change one side, you must make the identical change to the other, or the scale tips. Keep it balanced while you strip away everything sitting next to the variable. Strip it all away, keeping the scale level, and the variable is left alone with its value on the other side. That single idea solves every equation in this lesson.

Learning goals

Say what solving means, and confirm a solution by substituting
Apply the balance principle, doing the same to both sides
Undo what is attached with the inverse operation
Solve all four one-step forms, including $\dfrac{x}{a} = b$
Carry the method across negatives, fractions and decimals unchanged

What it means to solve an equation

To solve an equation is to find every value of the variable that makes the equation a true statement. Each such value is a solution. The equations in this lesson have exactly one solution apiece, so solving means pinning down that one number.

The honest way to test a candidate is to put it back into the original equation and check whether both sides come out equal. This is just the substitution you learned for evaluating expressions, now used as a verdict. Take $x + 5 = 12$ and try $x = 7$ :

7 + 5 = 12, \qquad 12 = 12. \;\checkmark

Both sides agree, so $7$ is a solution. Try instead $x = 6$ :

6 + 5 = 12, \qquad 11 = 12. \;\text{false}

The sides disagree, so $6$ is not a solution. Substitution never lies: a value is a solution exactly when it makes the two sides equal. You can always confirm an answer this way once you have it. What you need is a reliable way to produce the answer instead of guessing it, and that is the balance principle.

The balance principle

Think of the equals sign as the center of a balance scale, with each side of the equation in one pan. Saying the equation is true is the same as saying the two pans balance: they hold equal amounts. Now ask what you are allowed to do without disturbing that balance.

If two pans weigh the same and you add the same weight to both, they still weigh the same. If you remove the same weight from both, they still match. The same goes for doubling both, or halving both. Add or subtract the same amount on both sides, or multiply or divide both sides by the same nonzero number, and the equation stays true. This is the balance principle, and it is the one rule that powers every step of equation solving.

An equation is a balanced scale. The two pans hold equal amounts, so the same change made to both pans keeps them level. Change only one pan and the balance is lost.

Why does this preserve the solution? Because the operation is applied equally, a number that made the two sides equal before still makes them equal after.

Inverse operations undo what is attached to the variable

The balance principle says you may do the same thing to both sides. It does not yet say what to do. The goal tells you that: get the variable by itself on one side, a state called isolating the variable. Isolation is the goal because once $x$ stands alone the other side displays its value.

To isolate the variable you peel off whatever is attached to it, and you peel it off with the operation that undoes it. Two operations undo each other when doing one and then the other lands you back where you started. These are inverse operations:

Addition and subtraction are inverses. Adding $5$ then subtracting $5$ returns the original number.
Multiplication and division are inverses. Multiplying by $3$ then dividing by $3$ returns the original number.

So the plan for a one-step equation is short. Look at what is being done to the variable, then do the inverse to both sides. If $5$ is added to the variable, subtract $5$ from both sides. If the variable is multiplied by $3$ , divide both sides by $3$ . The inverse cancels the attachment on the variable’s side, leaving the variable alone, while the balance principle keeps the equation true. The one-step equations in this lesson all take one of four forms, one for each operation, and the rest of the lesson works through all four.

Both halves of that idea, the hunt for the number and the undoing, are things you can carry out on the number line below. It takes a number to start from and a number to add, and you set each one. The arrow draws the step and the filled point shows where it lands.

Try the hunt first. Leave the number added at $3$ and move the starting number along until the landing point reads $7$ . The only start that gets there is $4$ , and that hunt is the whole of $x + 3 = 7$ . You were looking for the number which, after a step of $3$ , arrives at $7$ , and the tool agrees: $x + 3 = 7$ , solved by $x = 4$ .

Now watch the undoing. Set the start to $7$ and the number added to $-3$ . You land on $4$ , and the tool now reads this walk as its own: $x + (-3) = 4$ , solved by $x = 7$ . Read the two readouts together. They say that stepping by $3$ and then stepping by $-3$ always lands you back where you began, whichever number you call the start. That is the cancellation that makes “subtract $3$ from both sides” the move that undoes “add $3$ ” in the equation you are solving.

Why the step back is what solves the equation

Start at Then add

2 + 3 = 5. That is the equation x + 3 = 5, solved by x = 2. Stepping back by (-3) undoes it and returns to 2.

One step on a number line, with both halves under your control: the number the step starts from, and the number added to it.

Form 1: the variable plus a number

An equation like $x + a = b$ has a number added to the variable. The inverse of adding $a$ is subtracting $a$ , so subtract $a$ from both sides.

Worked example 1 Solve $x + 8 = 21$

The variable has $8$ added to it. To undo that addition, subtract $8$ from both sides, keeping the scale balanced:

x + 8 - 8 = 21 - 8.

On the left, $8 - 8 = 0$ , so the $8$ is gone and the variable stands alone. On the right, $21 - 8 = 13$ :

x = 13.

Now check by substituting $13$ back into the original equation:

13 + 8 = 21, \qquad 21 = 21. \;\checkmark

Both sides match, so $x = 13$ is correct. Subtracting the same $8$ from each side is exactly the balance principle at work.

The numbers need not be whole. The method is identical with a decimal, a fraction, or a negative; you just carry out the subtraction using the arithmetic from earlier chapters.

Worked example 2 Solve $x + 2.5 = 4.1$

A decimal is added to the variable, so subtract that decimal from both sides:

x + 2.5 - 2.5 = 4.1 - 2.5.

The left side collapses to $x$ , and the right side is a decimal subtraction:

x = 1.6.

Check it against the original equation:

1.6 + 2.5 = 4.1, \qquad 4.1 = 4.1. \;\checkmark

The decimal changes nothing about the method: you still undo “add $2.5$ ” with “subtract $2.5$ .”

Check your understanding

Solve $x + 6 = 4$ for $x$ .

x = 2

x = -2

x = 10

x = -10

Form 2: the variable minus a number

An equation like $x - a = b$ has a number subtracted from the variable. The inverse of subtracting $a$ is adding $a$ , so add $a$ to both sides.

Worked example 3 Solve $x - 7 = 12$

Here $7$ has been subtracted from the variable. Undo it by adding $7$ to both sides:

x - 7 + 7 = 12 + 7.

On the left, $-7 + 7 = 0$ , so the variable is isolated. On the right, $12 + 7 = 19$ :

x = 19.

Check by substituting into the original equation:

19 - 7 = 12, \qquad 12 = 12. \;\checkmark

So $x = 19$ . Notice the symmetry with the previous form: a subtracted number is removed by adding it back. In the same way, an added number is removed by subtracting it.

Worked example 4 Solve $x - \tfrac{1}{4} = \tfrac{3}{4}$

A fraction has been subtracted from the variable, so add that fraction to both sides:

x - \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{3}{4} + \tfrac{1}{4}.

The left side becomes $x$ . On the right, the two fractions already share the denominator $4$ , so add the numerators:

x = \tfrac{3}{4} + \tfrac{1}{4} = \tfrac{4}{4} = 1.

Check the result in the original equation:

1 - \tfrac{1}{4} = \tfrac{3}{4}, \qquad \tfrac{3}{4} = \tfrac{3}{4}. \;\checkmark

So $x = 1$ . Adding fractions with a common denominator is the chapter-4 skill; the equation step is the same “add to both sides” as always.

Check your understanding

Which operation, applied to both sides, isolates the variable in $x - 9 = 15$ ?

A subtract

9

B add

9

C divide by

9

D multiply by

9

Form 3: a number times the variable

An equation like $ax = b$ has the variable multiplied by a number, the coefficient $a$ . The inverse of multiplying by $a$ is dividing by $a$ , so divide both sides by $a$ . You can always do this because the coefficient $a$ is a nonzero number. That matters because dividing by zero is undefined, as you saw in the division lessons.

Worked example 5 Solve $6x = 42$

The coefficient is $6$ , so the variable is multiplied by $6$ . Undo that by dividing both sides by $6$ :

\frac{6x}{6} = \frac{42}{6}.

On the left, dividing $6x$ by $6$ cancels the coefficient and leaves $x$ alone, since $6 \div 6 = 1$ and $1x = x$ . On the right, $42 \div 6 = 7$ :

x = 7.

Check by substituting into the original equation, writing the value in parentheses so the multiplication is clear:

6(7) = 42, \qquad 42 = 42. \;\checkmark

So $x = 7$ . Dividing by the coefficient is how you strip a multiplier off the variable.

A negative coefficient is handled the same way: divide both sides by the whole coefficient, sign and all. Then let the integer division rules from chapter 2 settle the sign.

Worked example 6 Solve $-4x = 20$

The coefficient is $-4$ , so divide both sides by $-4$ , carrying the negative sign with it:

\frac{-4x}{-4} = \frac{20}{-4}.

On the left, $-4$ divided by $-4$ is $1$ , leaving $x$ alone. On the right, a positive divided by a negative is negative:

x = \frac{20}{-4} = -5.

Check in the original equation:

-4(-5) = 20, \qquad 20 = 20. \;\checkmark

So $x = -5$ . The negative times negative on the left of the check is positive $20$ , which confirms the answer. Always divide by the full coefficient, including its sign.

Check your understanding

Solve $7x = 56$ for $x$ .

x = 8

x = 49

x = 63

x = 7

Form 4: the variable divided by a number

An equation like $\dfrac{x}{a} = b$ has the variable divided by a number. Here $a$ is nonzero, since dividing by zero is undefined and $\dfrac{x}{a}$ would not even make sense otherwise. The inverse of dividing by $a$ is multiplying by $a$ , so multiply both sides by $a$ .

Worked example 7 Solve $\dfrac{x}{3} = 9$

The variable is divided by $3$ . Undo that division by multiplying both sides by $3$ :

3 \cdot \frac{x}{3} = 3 \cdot 9.

On the left, multiplying by $3$ cancels the division by $3$ , since $\frac{3}{3} = 1$ , leaving the variable alone. On the right, $3 \cdot 9 = 27$ :

x = 27.

Check by substituting into the original equation:

\frac{27}{3} = 9, \qquad 9 = 9. \;\checkmark

So $x = 27$ . Multiplying by the divisor is the move that clears a fraction off the variable.

Worked example 8 Solve $\dfrac{x}{5} = -1.2$

The variable is divided by $5$ , so multiply both sides by $5$ , even though the right side is a negative decimal:

5 \cdot \frac{x}{5} = 5 \cdot (-1.2).

The left side becomes $x$ . On the right, $5 \cdot (-1.2) = -6$ , a positive times a negative:

x = -6.

Check it in the original equation:

\frac{-6}{5} = -1.2, \qquad -1.2 = -1.2. \;\checkmark

So $x = -6$ .

Check your understanding

Solve $\dfrac{x}{4} = 8$ for $x$ .

x = 2

x = 12

x = 32

x = 4

Reading the equation before you move

Every equation in this lesson comes down to the same two questions. First, what one operation is attached to the variable? Second, what is its inverse? Apply that inverse to both sides and the variable stands alone. This little table is the whole lesson in one glance.

If the equation reads	the variable is being	so apply to both sides	which gives
$x + a = b$	added to	subtract $a$	$x = b - a$
$x - a = b$	subtracted from	add $a$	$x = b + a$
$ax = b$	multiplied	divide by $a$	$x = \dfrac{b}{a}$
$\dfrac{x}{a} = b$	divided	multiply by $a$	$x = ab$

In the last two rows, $a$ is nonzero: you cannot divide by $a$ if $a$ is $0$ , and $\dfrac{x}{a}$ is not even defined when $a$ is $0$ .

The pattern to internalize is that you always apply the inverse operation. Adding is undone by subtracting, multiplying by dividing, and the reverse each way. Whatever you choose, you do it to both sides, and you finish by substituting your answer back to confirm both sides are equal.

Worked example 9 Identify the move, then solve $\tfrac{2}{3}x = 8$

First read the equation. The variable is multiplied by the coefficient $\tfrac{2}{3}$ , so this is the multiplication form, and the inverse of multiplying by $\tfrac{2}{3}$ is dividing by $\tfrac{2}{3}$ . Dividing by a fraction is the same as multiplying by its reciprocal, the chapter-4 rule, so multiply both sides by $\tfrac{3}{2}$ :

\frac{3}{2} \cdot \frac{2}{3}x = \frac{3}{2} \cdot 8.

On the left, $\tfrac{3}{2} \cdot \tfrac{2}{3} = 1$ , leaving the variable alone. On the right, $\tfrac{3}{2} \cdot 8 = \tfrac{24}{2} = 12$ :

x = 12.

Check in the original equation:

\frac{2}{3}(12) = \frac{24}{3} = 8, \qquad 8 = 8. \;\checkmark

So $x = 12$ . Reading the form first told you to multiply by the reciprocal, and the check confirms it.

Check your understanding

You solve $x + 4 = 11$ and get $x = 7$ . Substituting to check, what should the left side equal?

7

11

4

15

Common mistakes

Changing only one side. If you subtract $5$ from the left, you must subtract $5$ from the right too. Touch one pan and the scale tips, so the new equation no longer has the same solution.
Using the same operation instead of the inverse. To undo $x + 5$ you subtract $5$ , not add another $5$ . Adding $5$ to both sides of $x + 5 = 12$ gives $x + 10 = 17$ , which is no closer to isolating the variable.
Subtracting a coefficient instead of dividing. In $6x = 42$ the $6$ is multiplied by the variable, so you divide both sides by $6$ . Subtracting $6$ does nothing useful, because $6x$ is not $6 + x$ .
Dividing when you should multiply. In $\dfrac{x}{3} = 9$ the variable is already divided by $3$ , so you multiply both sides by $3$ . Dividing again would only shrink the variable further.
Losing a negative sign. When the coefficient is $-4$ , divide by $-4$ , sign included, so the sign of the answer comes out right. Dividing by a bare $4$ drops the negative.
Skipping the check. Substituting your answer back into the original equation costs one line and catches almost every arithmetic slip. A solution that does not make both sides equal is wrong, full stop.

Key takeaways

To solve an equation is to find the value of the variable that makes it true. That value is a solution.
The balance principle: an equation is a balanced scale. Add or subtract the same amount on both sides, or multiply or divide both sides by the same nonzero number, and every solution is kept.
Isolate the variable by undoing what is attached to it with the inverse operation. Addition and subtraction undo each other, and multiplication and division undo each other.
The four one-step forms, with $a$ nonzero: $x + a = b$ subtract $a$ ; $x - a = b$ add $a$ ; $ax = b$ divide by $a$ ; $\dfrac{x}{a} = b$ multiply by $a$ .
The method does not change for negative, fraction, or decimal values; you simply carry out the inverse operation with the arithmetic from earlier chapters.
Always check by substituting your answer into the original equation. That is what confirms a solution is correct.

Solving One-Step Equations

Learning goals

What it means to solve an equation

The balance principle

Inverse operations undo what is attached to the variable

Form 1: the variable plus a number

Form 2: the variable minus a number

Form 3: a number times the variable

Form 4: the variable divided by a number

Reading the equation before you move

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice