Why do I undo addition before multiplication when solving?

Because you are reversing the order of operations. To build 4x - 7 from x, you multiply by 4 first and subtract 7 last, so to take it apart you remove the last thing first. Dividing first is still legal, but then you must divide every term on both sides, which usually creates fractions you did not need.

What do I do when the variable has a fraction in front of it?

Multiply both sides by the reciprocal of that fraction. For 4/7x = 8, multiply both sides by 7/4 to get x = 14. Alternatively you can multiply both sides by the denominator and then divide by the numerator, which gives the same answer in two steps instead of one.

Do I really have to check every solution?

Checking takes about fifteen seconds and catches the sign errors and arithmetic slips that cause most wrong answers in this unit. It also gives you an independent confirmation, so you are not just trusting the steps you already wrote. Always substitute into the original equation, since substituting into a line you rewrote would hide an error made earlier.

What if the variable ends up on the right side, like 12 = x + 5?

Nothing changes. Equality works both ways, so you can solve it in place — subtract 5 from both sides to get 7 = x — or rewrite it as x + 5 = 12 first. Either way the solution is x = 7. Getting comfortable with the variable on either side prepares you for equations with variables on both sides in the next lesson.

ALG1-2.1

One- & Two-Step Equations

Learn to solve one- and two-step equations with inverse operations — negative coefficients, fractions, and checking every solution by substitution.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on One- & Two-Step Equations, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start One- & Two-Step Equations free

What this lesson covers

An equation is a claim that two expressions name the same number. Solving it means finding every value of the variable that makes the claim true, and the tool for that job is the inverse operation: addition undoes subtraction, multiplication undoes division. Everything you do in later units — multi-step equations, rearranging formulas, proportions, systems — is built on the moves you practice here.

In this lesson you will isolate the variable in equations like

x - 9 = -4

-6y = 42

, and

\frac{2}{3}x - 7 = -3

, keeping the equation balanced at every step. You will also learn to justify each step (why is the new equation equivalent to the old one?) and to verify your answer by substituting it back into the original equation. That check is not busywork — it is how you catch a dropped negative sign before it follows you through a whole page of homework.

What "Solving" Actually Means: Balance and Equivalence

The equals sign is a balance point, not a signal to "do something." In

3x + 4 = 19

, the expression

3x + 4

and the number

19

are two names for the same value. Solving means rewriting the equation, over and over, into simpler equivalent equations — equations with exactly the same solution — until the variable stands alone.

What guarantees equivalence? The properties of equality. If two quantities are equal and you add the same number to both, they stay equal. Same for subtracting, multiplying, or dividing both sides by the same nonzero number. That last word matters: dividing both sides by zero is never allowed, which is why you can never "divide by

x

" when

x

might be zero.

So every legal step has the same shape: apply one operation to both sides. Writing

3x + 4 = 19

and then

3x = 15

is shorthand for subtracting

4

from each side.

A solution is a number that makes the original sentence true. Substituting

x = 5

into

3x + 4 = 19

gives

19 = 19

, a true statement, so

5

is the solution. Substituting

x = 4

gives

16 = 19

, false, so

4

is not. This is the whole logic of the unit: legal steps preserve truth, and substitution confirms it.

Where students go wrong early is treating the equals sign as "here comes the answer" and operating on only one side. Once one side changes and the other does not, the balance is broken and every line after it is a different problem.

One-Step Equations and Their Inverses

A one-step equation has exactly one operation attached to the variable. Identify that operation, then apply its inverse to both sides.

Equation	What is done to the variable	Inverse step	Solution
$x + 12 = 5$	add $12$	subtract $12$	$x = -7$
$n - 8 = -3$	subtract $8$	add $8$	$n = 5$
$-6y = 42$	multiply by $-6$	divide by $-6$	$y = -7$
$\frac{m}{4} = -3$	divide by $4$	multiply by $4$	$m = -12$
$\frac{3}{5}t = 9$	multiply by $\frac{3}{5}$	multiply by $\frac{5}{3}$	$t = 15$

Two cases deserve extra attention. First, a fractional coefficient: rather than dividing by

\frac{3}{5}

, multiply both sides by the reciprocal

\frac{5}{3}

, since

\frac{5}{3}\cdot\frac{3}{5} = 1

. Multiplying by the reciprocal and dividing by the fraction give the same result, but the reciprocal method is faster and less error-prone.

Second, the bare negative:

-x = 7

is not solved. The coefficient is

-1

, so divide both sides by

-1

(or multiply by

-1

) to get

x = -7

. Students routinely leave

-x = 7

as if it were finished, or write

x = 7

.

Also remember that subtraction is addition of a negative. In

x - 8 = -3

, the number attached is

-8

, so you add

8

x = -3 + 8 = 5

. Reading the sign as part of the number prevents most sign errors before they happen.

Two-Step Equations: Undo in Reverse Order

A two-step equation such as

4x - 7 = 13

has two operations stacked on the variable: multiply by

4

, then subtract

7

. To unwrap it, undo in the reverse of the order of operations — deal with addition and subtraction first, then multiplication and division.

4x - 7 = 13

becomes

4x = 20

(add

7

to both sides), then

x = 5

(divide both sides by

4

).

Think of it as taking off shoes and socks in the opposite order you put them on. The constant is the outermost layer, so it comes off first.

Negative coefficients follow the same plan; just carry the sign carefully. For

8 - 5x = 23

, the variable term is

-5x

, so subtract

8

from both sides to get

-5x = 15

, then divide by

-5

to get

x = -3

. A very common wrong answer here is

x = 3

, from dividing by

5

instead of

-5

.

Fractional forms come in two flavors. When the fraction is a coefficient, as in

\frac{x}{3} + 6 = 10

, subtract

6

to get

\frac{x}{3} = 4

, then multiply by

3

to get

x = 12

. When the whole side is over a denominator, as in

\frac{x + 5}{2} = -1

, the division applies to the entire numerator, so multiply both sides by

2

first:

x + 5 = -2

, so

x = -7

. Mixing these two structures up — subtracting the

5

before clearing the denominator — is one of the most frequent mistakes in this lesson.

Verifying by Substitution and Catching Errors

Checking is the step that turns a guess into a justified answer. Substitute your solution into the original equation — not into a line you rewrote, since an error there would go undetected — and simplify each side separately until you see either a true statement or a false one.

Check

x = -3

8 - 5x = 23

: the left side is

8 - 5(-3) = 8 + 15 = 23

, and the right side is

23

. Since

23 = 23

, the solution is verified. Notice how the parentheses around

-3

keep the double negative visible; writing

8 - 5 \cdot -3

without them is where sign slips happen.

Three error patterns show up again and again:

Operating on one side only. If you add

7

to the left, add

7

to the right in the same line.

Undoing in the wrong order. Dividing

4x - 7 = 13

4

first is legal, but then you must divide every term:

x - \frac{7}{4} = \frac{13}{4}

. Students usually write

x - 7 = \frac{13}{4}

, which is a different equation.

Losing a negative on the coefficient. After reaching

-5x = 15

, dividing by positive

5

gives

-x = 3

, and stopping there leaves the problem unfinished.

A complete written solution shows the original equation, one step per line with the inverse operation applied to both sides, a boxed or clearly stated answer, and a substitution check. That format makes your reasoning visible and makes errors easy for you to find yourself.

Key terms

Equation.: A statement that two expressions are equal, such as $2x + 1 = 9$ . It is true for some values of the variable and false for others.
Solution of an equation.: A value of the variable that makes the equation a true statement when substituted in.
Inverse operation.: An operation that undoes another: addition and subtraction are inverses, as are multiplication and division (by a nonzero number).
Equivalent equations.: Equations with exactly the same solution set. Applying the same legal operation to both sides produces an equivalent equation.
Properties of equality.: The rules that let you add, subtract, multiply, or divide both sides of an equation by the same value (nonzero for division) without changing its solution.
Coefficient.: The number multiplying the variable. In $-5x$ the coefficient is $-5$ ; in $-x$ it is $-1$ .
Reciprocal.: The multiplicative inverse of a nonzero number: the reciprocal of $\frac{3}{5}$ is $\frac{5}{3}$ , and their product is $1$ . Multiplying by the reciprocal clears a fractional coefficient.
Substitution check.: Replacing the variable in the original equation with your solution and simplifying both sides to confirm they are equal.

Worked example

Solve

\frac{2}{3}x - 7 = -3

and verify the solution by substitution.

Identify what is being done to

x

. First

x

is multiplied by

\frac{2}{3}

, then

7

is subtracted. Undo in reverse order, so handle the subtraction first.

Step 1: Add

7

to both sides.

\frac{2}{3}x - 7 + 7 = -3 + 7

\frac{2}{3}x = 4

Step 2: The coefficient is the fraction

\frac{2}{3}

, so multiply both sides by its reciprocal

\frac{3}{2}

\frac{3}{2}\cdot\frac{2}{3}x = \frac{3}{2}\cdot 4

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

Step 3: Verify in the original equation. Substitute

x = 6

into

\frac{2}{3}x - 7

\frac{2}{3}(6) - 7 = 4 - 7 = -3

The right side is

-3

, and

-3 = -3

is true, so

x = 6

is the solution.

Notice two things. Adding

7

before touching the fraction avoided any messy arithmetic with fractional constants. And multiplying by

\frac{3}{2}

was cleaner than dividing

4

\frac{2}{3}

, even though both give

6

Practice questions

Solve

5 - 3x = 20

$x = -5$
$x = 5$
$x = -\frac{25}{3}$
$x = \frac{25}{3}$

Answer: $x = -5$

The variable term is

-3x

, and the constant on that side is

+5

. Subtract

5

from both sides:

-3x = 15

. Now divide both sides by

-3

, not by

3

x = -5

. The choice

x = 5

comes from dividing by positive

3

and dropping the negative, and the choices with

25

come from adding

5

instead of subtracting it. Check:

5 - 3(-5) = 5 + 15 = 20

, which matches the right side.

A student solves

-x + 9 = 4

and writes the answer

x = -5

. Identify the error, give the correct solution, and show the substitution check.

Answer: The student stopped at $-x = -5$ and reported that as the value of $x$ ; the correct solution is $x = 5$ .

Subtracting

9

from both sides gives

-x = -5

, which is a true intermediate step. But

-x

is not

x

— the coefficient is

-1

, so the variable is not yet isolated. Divide both sides by

-1

(or multiply by

-1

) to get

x = 5

. Substituting into the original equation:

-(5) + 9 = -5 + 9 = 4

, which matches the right side, so

x = 5

checks out. Substituting the student's

-5

would give

-(-5) + 9 = 14

, not

4

, which is exactly how the substitution check catches this error.

A repair shop charges a flat fee of 45 dollars plus 30 dollars for each hour of labor. A customer's bill was 165 dollars. Write and solve a two-step equation to find the number of labor hours, then verify your answer.

Answer: $30h + 45 = 165$ , so $h = 4$ hours.

Let

h

be the number of labor hours. The labor charge is

30h

and the flat fee is added once, giving the equation

30h + 45 = 165

. Subtract

45

from both sides to undo the constant:

30h = 120

. Then divide both sides by

30

h = 4

. Verify in the original equation:

30(4) + 45 = 120 + 45 = 165

, which matches the bill. Setting up the equation is the part students most often rush — the flat fee is added once, not multiplied by the hours, so it belongs outside the

30h

term.

FAQ

Why do I undo addition before multiplication when solving?: Because you are reversing the order of operations. To build $4x - 7$ from $x$ , you multiply by $4$ first and subtract $7$ last, so to take it apart you remove the last thing first. Dividing first is still legal, but then you must divide every term on both sides, which usually creates fractions you did not need.
What do I do when the variable has a fraction in front of it?: Multiply both sides by the reciprocal of that fraction. For $\frac{4}{7}x = 8$ , multiply both sides by $\frac{7}{4}$ to get $x = 14$ . Alternatively you can multiply both sides by the denominator and then divide by the numerator, which gives the same answer in two steps instead of one.
Do I really have to check every solution?: Checking takes about fifteen seconds and catches the sign errors and arithmetic slips that cause most wrong answers in this unit. It also gives you an independent confirmation, so you are not just trusting the steps you already wrote. Always substitute into the original equation, since substituting into a line you rewrote would hide an error made earlier.
What if the variable ends up on the right side, like $12 = x + 5$ ?: Nothing changes. Equality works both ways, so you can solve it in place — subtract $5$ from both sides to get $7 = x$ — or rewrite it as $x + 5 = 12$ first. Either way the solution is $x = 7$ . Getting comfortable with the variable on either side prepares you for equations with variables on both sides in the next lesson.

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The Crimsora tutor teaches One- & Two-Step Equations live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

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