Solving linear equations with one variable

Linear equations in one variable are equations where the variable has an exponent of 1, which is typically not shown (it is understood). An example would be something like \(12x = x – 5\). To solve linear equations, there is one main goal: isolate the variable. In this lesson, we will look at how this is done through several examples.

Table of Contents

  1. Examples of solving one-step equations
  2. Examples of solving two-step equations
  3. Examples of equations where you must simplify first
  4. Infinitely many or no solutions
  5. Summary


Examples of solving one-step linear equations

After all your hard work solving the equation, you know that you want a final answer like \(x=5\) or \(y=1\). In both of these cases the variable is isolated, or by itself.

So we need to figure out how to isolate the variable. How we do this depends on the equation itself! If it was multiplied by something, we will divide. If something was added to it, we will subtract. By doing this, we will slowly be getting the variable by itself.

Let’s use an example to see how this works.


Solve the equation: \(4x = 8\)


In this example, the 4 is multiplying the \(x\). Therefore, to isolate \(x\), you must divide that side by 4. When doing this, you must remember one important rule: whatever you do to one side of the equation, you must do to the other side. So we will divide both sides by 4.

\(\begin{align}4x &= 8 \\ \dfrac{4x}{\color{red}{4}} &= \dfrac{8}{\color{red}{4}}\end{align}\)


\(x = \boxed{2}\)

That’s it, one step and we are done. (That’s why equations like these are often called “one-step” equations)


Anytime you are solving linear equations, you can always check your answer by substituting it back into the equation. If you get a true statement, then the answer is correct. This isn’t 100% necessary for every problem, but it is a good habit so we will do it for our equations.

In this example, our original equation was \(4x = 8\). To check this, verify the following is true:

\(\begin{align}4x &= 8\\ 4(2) &= 8 \\ 8 &= 8\end{align}\)

This is a true statement, so our answer is correct.

For any equation, whatever operation you do to one side must also be done to the other side

Let’s try a couple more examples before moving on to more complex equations.


Solve: \(3x=12\)


Since \(x\) is being multiplied by 3, the plan is to divide by 3 on both sides:

\(\begin{align}3x &=12\\ \dfrac{3x}{\color{red}{3}} &=\dfrac{12}{\color{red}{3}}\\ x&= \boxed{4}\end{align}\)


To check our answer, we will let \(x = 4\) and substitute it back into the equation:

\(\begin{align}3x &= 12\\3(4) &= 12 \\ 12 &= 12\end{align}\)

Just as before, since this is a true statement, we know our answer is correct.

In the next example, instead of the variable being multiplied by a value, a value is being subtracted from the variable. To “undo” this, we will add that value to both sides.


Solve: \(y-9=21\)


This time, 9 is being subtracted from y. So, we will undo that by adding 9 to both sides.

\(\begin{align}y-9&=21\\ y-9 \color{red}{+9}&=21\color{red}{+9}\\y&=30\end{align}\)

Next we will look at what are commonly called “two-step” equations. In these equations, we will need to undo two operations in order to isolate the variable.

Examples of two step equations

In each of the examples above, there was a single step to perform before we had our answer. In these next examples, you will see how to work with equations that have two steps instead. If there is more than one operation, it is important to remember the order of operations, PEMDAS. Since you are undoing the operations to \(x\), you will work from the “outside in”. This is easier to understand when you see it in an example.


Solve: \(2x-7=13\)


Notice the two operations happening to \(x\): it is being multiplied by 2 and then having 7 subtracted. We will need to undo these. But, only the \(x\) is being multiplied by 2, so the first step will be to add 7 to both sides. Then we can divide both sides by 2.

Adding 7 to both sides:

\(\begin{align} 2x-7 &= 13\\ 2x-7 \color{red}{+7} & =13 \color{red}{+7}\\ 2x&=20\end{align}\)

Now divide both sides by 2:

\(\begin{align} 2x &=20 \\ \dfrac{2x}{\color{red}{2}}&=\dfrac{20}{\color{red}{2}}\\ x&= \boxed{10}\end{align}\)


Just like with simpler problems, you can check your answer by substituting your value of \(x\) back into the original equation.

\(\begin{align}2x-7&=13\\ 2(10) – 7 &= 13\\ 13 &= 13\end{align}\)

This is true, so we have the correct answer.

Let’s look at one more two-step example before we jump up in difficulty again. Make sure that you understand each step shown and work through the problem as well.


Solve: \(5w + 2 = 9\)


As above, there are two operations: \(w\) is being multiplied by 5 and then has 2 added to it. We will undo these by first subtracting 2 from both sides and then dividing by 5.

\(\begin{align}5w + 2 &= 9\\ 5w + 2 \color{red}{-2} &= 9 \color{red}{-2}\\ 5w &= 7\\ \dfrac{5w}{\color{red}{5}} &=\dfrac{7}{\color{red}{5}}\\w=\boxed{\dfrac{7}{5}}\end{align}\)

The fraction on the right can’t be simplified, so that is our final answer.


Let \(w = \dfrac{7}{5}\). Then:

\(\begin{align}5w + 2 &= 9\\ 5\left(\dfrac{7}{5}\right) + 2 &= 9\\ 7 + 2 &= 9\\ 9 &= 9 \end{align}\)

So, we have the correct answer once again!

Simplifying before solving

In the following examples, there are more variable terms and possibly some simplification that needs to take place. In each case, the steps will be to first simplify both sides, then use what we have been doing to isolate the variable. We will first take an in depth look at an example to see how this all works.

To understand this section, you should be comfortable with combining like terms.


Solve: \(3x+2=4x-1\)


Since both sides are simplified (there are no parentheses we need to figure out and no like terms to combine), the next step is to get all of the x’s on one side of the equation and all the numbers on the other side. The same rule applies – whatever you do to one side of the equation, you must do to the other side as well!

It is possible to either move the \(3x\) or the \(4x\). Suppose you moved the \(4x\). Since it is positive, you would do this by subtracting it from both sides:

\(\begin{align}3x+2 &=4x-1\\ 3x+2\color{red}{-4x} &=4x-1\color{red}{-4x}\\ -x+2 & =-1\end{align}\)

Now the equation looks like those that were worked before. The next step is to subtract 2 from both sides:

\(\begin{align}-x+2\color{red}{-2} &= -1\color{red}{-2}\\-x=-3\end{align}\)

Finally, since \(-x= -1x\) (this is always true), divide both sides by \(-1\):

\(\begin{align}\dfrac{-x}{\color{red}{-1}} &=\dfrac{-3}{\color{red}{-1}}\\ x&=3\end{align}\)


You should take a moment and verify that the following is a true statement:

\(3(3)+ 2 = 4(3) – 1\)

In the next example, we will need to use the distributive property before solving. It is easy to make a mistake here, so make sure that you distribute the number in front of the parentheses to all the terms inside.


Solve: \(3(x+2)-1=x-3(x+1)\)


First, distribute the 3 and –3, and collect like terms.

\(\begin{align} 3(x+2)-1 &=x-3(x+1)\\ 3x+6-1&=x-3x-3 \\ 3x+5&=-2x-3\end{align}\)

Now we can add 2x to both sides. (Remember you will get the same answer if you instead subtracted 3x from both sides)

\(\begin{align} 3x+5\color{red}{+2x} &=-2x-3\color{red}{+2x}\\ 5x+5& =-3\end{align}\)

From here, we can solve as we did with other two-step equations.

\(\begin{align}5x+5\color{red}{-5} &=-3\color{red}{-5}\\ 5x &=-8\\ \dfrac{5x}{\color{red}{5}}&=\dfrac{-8}{\color{red}{5}}\\ x &= \dfrac{-8}{5} \\ &=\boxed{-\dfrac{8}{5}}\end{align}\)


This was a tough one, so remember to check your answer and make sure no mistake was made. To do that, you will be making sure that the following is a true statement:


(Note: it does work – but you have to be really careful about parentheses!)

Infinitely many solutions and no solutions

There are times when you follow all of these steps and a really strange solution comes up. For example, when solving the equation \(x+2=x+2\) using the steps above, end up with \(0=0\). This is certainly true but what good does it do?

If you get a statement such as this, it means that the equation has infinitely many solutions. Any \(x\) you could think of would satisfy the equation \(x+2=x+2\). The appropriate answer in this case is “infinitely many solutions”.

The other situation comes up when you simplify an equation down into a statement that is never true such as \(3=4\) or \(0=1\). This happens with the equation \(x+5=x-7\) which will lead to \(5= -7\), something that is certainly never true. This means that no \(x\) would satisfy this equation. In other words “no solution”. In summary:

  • If you get a statement that is always true like \(5 = 5\) or \(0 = 0\), then there are infinitely many solutions.
  • If you get a statement that is always false like \(10 = 11\) or \(1 = 5\), then there are no solutions.



Solving linear equations is all about isolating the variable. Depending on the equation, this may take as little as one step or many more steps. Always check if you need to simplify one or both sides of the equation first, and always check your answer.