Solving Two-Step Linear Equations Using Inverse Operations

Hello, and welcome to this first lesson on solving systems by substitution. Before substitution can work smoothly, you need to be able to solve the one-variable equations it creates. This module builds that essential skill: isolating a variable, then later using its value in a system and checking the final ordered pair.

Today, you will solve two-step equations in the form

$$ax+b=c$$

using inverse operations while preserving equality. By the end, you should be able to show clear algebraic steps and verify that your answer makes the original equation true.

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An equation is a balanced statement

An equation says that its left and right sides have the same value. For example,

$$6x+2=26$$

means that whatever $6x+2$ equals is exactly the same as $26$. Your job is to find the number $x$ that makes this statement true.

Think of the expression on the variable side as a sequence of operations:

1. Start with $x$.
2. Multiply it by $6$.
3. Add $2$.

To get $x$ alone, undo those operations in reverse order:

1. Undo the addition or subtraction first.
2. Undo the multiplication or division second.

Every operation must be performed on both sides of the equation. Otherwise, the equality is no longer balanced.

The phrase inverse operation simply means an operation that undoes another:

Operation affecting the variable	Inverse operation
Add $b$	Subtract $b$
Subtract $b$	Add $b$
Multiply by $a$	Divide by $a$
Divide by $a$	Multiply by $a$

For the general form

$$ax+b=c$$

the constant $b$ is added last, so remove it first:

$$ax+b-b=c-b$$ $$ax=c-b$$

Then divide by the coefficient $a$, assuming $a\ne 0$:

$$\frac{ax}{a}=\frac{c-b}{a}$$ $$x=\frac{c-b}{a}$$

You do not need to memorize that final formula. The reliable method is to identify what is being done to the variable and undo it one layer at a time.

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A complete worked example

Consider:

$$6x+2=26$$

The $+2$ is outside the multiplication by $6$, so it is the first operation to undo. Subtract $2$ from both sides:

$$6x+2-2=26-2$$ $$6x=24$$

Now $x$ is still being multiplied by $6$. Divide both sides by $6$:

$$\frac{6x}{6}=\frac{24}{6}$$ $$x=4$$

A solution should be checked in the original equation:

$$6(4)+2=26$$ $$24+2=26$$

Since the statement is true, $x=4$ is the solution.

The balance image is useful because it emphasizes a key point: subtraction, addition, multiplication, and division are not things you “move across” an equal sign. Instead, you apply the same operation to both complete sides.

For example, in

$$4y-5=15$$

the $-5$ is undone by adding $5$ to both sides. After that, the coefficient $4$ is undone by dividing both sides by $4$. The operation is chosen from the sign and operation visible in the equation, not from a memorized guess.

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Signs and coefficients: slow down at the second step

A coefficient can be negative. The method does not change; only the arithmetic demands care. Consider:

$$-3p+4=-11$$

First, remove the $+4$ by subtracting $4$ from both sides:

$$-3p+4-4=-11-4$$ $$-3p=-15$$

Next, divide both sides by $-3$:

$$\frac{-3p}{-3}=\frac{-15}{-3}$$ $$p=5$$

The answer is positive because a negative divided by a negative is positive. Check it:

$$-3(5)+4=-11$$ $$-15+4=-11$$

The same reasoning works when division is written explicitly. For instance, in

$$\frac{r}{5}+8=11$$

remove $8$ first, leaving

$$\frac{r}{5}=3$$

Then multiply both sides by $5$ to obtain $r$. Although the notation looks different, $\frac{r}{5}$ means $\frac{1}{5}r$: it is still a two-step equation of the form $ax+b=c$.

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Avoiding the most common errors

Most mistakes in two-step equations are not about difficult algebra. They come from losing track of an operation, a sign, or one side of the equation.

Mistake	Why it causes trouble	Reliable correction
Changing only one side	The two sides no longer have equal values.	Write the same operation on both sides every time.
Dividing before removing the constant	It can be legal if every term is divided, but it usually creates unnecessary fractions.	Undo addition or subtraction first.
Using the wrong inverse	For example, subtracting $5$ to undo $-5$.	Read the sign: undo $-5$ by adding $5$.
Stopping at $6x=24$	The variable is still multiplied by $6$.	Continue until the variable has coefficient $1$.
Checking the last simplified line only	Earlier work could contain an error.	Substitute into the original equation.

A clean written layout makes errors easier to notice. Keep each transformation on its own line:

$$2x-6=10$$ $$2x-6+6=10+6$$ $$2x=16$$ $$\frac{2x}{2}=\frac{16}{2}$$ $$x=8$$

The first step adds $6$, because $+6$ is the inverse of subtracting $6$. The second divides by $2$, because division by $2$ undoes multiplication by $2$.

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A compact decision routine

When you see an equation like

$$ax+b=c$$

use this routine:

1. Locate the constant added to or subtracted from the variable term.
2. Apply its inverse to both sides.
3. Simplify until only $ax$ remains on the variable side.
4. Divide by the coefficient $a$, on both sides.
5. Check by substituting your answer into the original equation.

In the next lesson, this procedure becomes part of substitution. Once one equation in a system gives you a one-variable equation, the same inverse-operation routine will let you solve it confidently.

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You have established the algebraic foundation for substitution: equations stay balanced when you do the same thing to both sides, and operations are undone in reverse order. For $ax+b=c$, remove the constant first, then remove the coefficient, and finally check the result in the original equation.

Next, you will substitute an isolated expression for one variable into a second equation and solve the resulting one-variable equation.