Solving 1-Variable Linear Equations

Before we look at solving equations, we will first define what exactly an equation.

Definition: An Equation is an equivalence of two values. (E.g. $2 = 1 + 1$ or $2x = 4 - x$) Each value in the equation is known as an Expression (E.g. $2x$ or $5 - x$).

Consider the equation $2x = 6$. We say that a solution to the equation is a value of $x$ that such that both sides of the equal sign are equal. By inspection we can see that $x = 3$ is a solution to the equation.

Of course, some equations are more complicated and inspection is not always the best method to find a solution. For example, consider the equation $2x + 14 = 18x - 4$. One way to find a solution to this equation is to move all expressions with variables to one side of the equation, and move all expressions without variables to the other side of the equation. For example:

(1)

\begin{align} 2x + (-2x) + 14 = 18x + (-2x) - 4 \quad \mathrm{[Subtracting \: 2x \: from \: both \: sides]} \\ 14 = 16x - 4 \\ 14 + (4) = 16x - 4 + (4) \quad \mathrm{[Adding \: 4 \: to \: both \: sides]} \\ 18 = 16x \end{align}

We now want to isolate the variable $x$. We can do so by dividing both sides of the equation by the number in front of $x$ to get:

(2)

\begin{align} \frac{18}{16} = \frac{16}{16}x \\ \frac{18}{16} = x \end{align}

Thus we have a solution $x = \frac{18}{16}$. Note that you can always check to see if your answer is correct by plugging it back into both sides of the expression and verifying both sides are equal. Let's verify with the left side first:

(3)

\begin{align} 2x + 14 = ? \\ 2(\frac{18}{16}) + 14 = 16.25 \\ \end{align}

Now let's verify with the right side of the equation:

(4)

\begin{align} 18x - 4 = ? \\ 18(\frac{18}{16}) - 4 = 16.25 \end{align}

Since both sides of the equation are equal, we know that our solution $x = \frac{18}{16}$ is correct. We will summarize the steps necessary to solve an equation of this form (1-variable linear equations).

Method for Finding Solutions to 1-Variable Linear Equations

Step 1	Simplify both sides of the equation. For example, if we have two expressions containing variables (or numbers), we should add them together, for example $2x + x = 4 +2$ should be simplified to $3x = 6$.
Step 2	Move all expressions with variables to one side of the equation by addition/subtraction. We note that the side chosen does not matter.
Step 3	Move all expressions without variables to the other side of the equation by addition/subtraction.
Step 4	If there is a number in front of the variable, divide both sides of the equation by that number to obtain the solution.

Example 1

Find a solution to the equation $4x - 16 = 3x + 9$.

We will follow the steps outlined in the method above.

Step 1	Both sides of the equation are simplified already.
Step 2	Let's first move the $x$'s to the left side of the equation by subtracting the expression $3x$ from both sides to get $x - 16 = 9$.
Step 3	Let's now move all non-variable expressions to the right side of the equation by adding $16$ to both sides of the equation to get $x = 25$.
Step 4	There is no number in front of our variable, and our solution is $x = 25$.

Of course it is important to verify our answer. We thus get that:

(5)

\begin{align} 4(25) - 16 = 84 \quad \mathrm{and} \quad 3(25) + 9 = 84 \end{align}

Exercise Questions

Solve the following equations and verify the solutions.

• 1. $2x - 8 = 4$

• 2. $6x = 4 - 3x$

• 3. $2x + 3 = 5x - 3$

• 4. $18x - 23 = 13x + 3$

• 5. $x + 2x + 3x = 6$.

• 6. $3x - 5 + 5x = 12$.

• 7. $2x - 9 + 2 = x + 7 - 3x$.

• 8. Is there ever a scenario when a 1-variable linear equation (like all of the examples on this page) will have no solutions or more than one solution? Explain.

Solutions

See solutions

• 1. $x = 6$

• 2. $x = \frac{4}{9}$.

• 3. $x = 2$.

• 4. $\frac{26}{5}$.

• 5. $x = 1$.

• 6. $x = \frac{17}{8}$.

• 7. $x = 4$.

• 8. No. A linear equation will always have 1 solution. We can always reduce a 1-variable linear equation to the form $x = a$, so $a$ must be the solution. There can only be one value for $a$, so there is always only one solution.