# Solving Linear Equations

The process of solving linear equations involves isolating the variable on one side of the equation and then finding its value. This can be done by using inverse operations such as addition, subtraction, multiplication, and division.

For example, to solve the equation $3x+5=11$ , we can subtract 5 from both sides to get

$3x+5-5=11-5$

$3x=6$

Then, we can divide both sides by 3 to get

$\frac{3x}{3}=\frac{6}{3}$

$x=2$

So the solution to the equation $3x+5=11$ is $x=2$ .

We can also check our answer by plugging it back into the original equation to make sure it makes a true statement.

$3\times 2+5=11$

$6+5=11$

Since this is a true statement, we have correctly solved the equation.

## Inverse operations

To start with, you need to understand inverse operations. The inverse of an operation is its opposite, or the operation that gets you back to the number you started with.

For example, if you started with the number 6, and then add 4:

$6+4=10$

To get back from 10 to 6, you have to subtract 4 from 10.

$10-4=6$

Therefore, addition and subtraction are inverse operations.

Similarly, division is the inverse operation of multiplication, and vice versa.

$7\times 5=35$

$\frac{35}{5}=7$

It's important to know your inverse operations when solving linear equations, as you will see in the next section.

## Properties of linear equations

- The value of the variable that makes the linear equation true is called the solution or root of the linear equation.
- The solution of a linear equation is unaffected if the same number is added, subtracted, multiplied, or divided into both sides of the equation.
- The graph of a linear equation in one or two variables always forms a straight line.

## One-step linear equations

Some linear equations can be solved with only a single operation. For this type of equation, you will use the inverse operation to solve.

**Example 1**

Solve for n.

$n+8=10$

The inverse operation of addition is subtraction. So you will subtract 8 from both sides.

$n+8-8=10-8$

$n=2$

**Example 2**

Solve for y.

$\frac{3}{4y}=15$

The inverse operation of multiplication is division. So you must divide both sides by 3/4, which is the same as multiplying by $\frac{4}{3}$ .

$\frac{4}{3}\times \frac{3}{4}y=\frac{4}{3}\times 15$

$y=20$

## Two-step linear equations

More commonly, we will need two operations to solve a linear equation.

**Example 3**

Solve for x.

$3x+5=11$

To isolate the variable, we follow the order of operations in reverse. We undo the addition before we undo the multiplication.

$3x+5-5=11-5$

Subtract 5 from both sides.

$3x=6$

We have undone one operation. One more to go.

$\frac{3x}{3}=\frac{6}{3}$

Divide each side by 3.

$x=2$

We have solved the equation!

The thing that makes the equation linear is that the highest power of x is ${x}^{1}$ (no ${x}^{2}$ or other powers-for those, see quadratic equations and polynomials).

## More complex linear equations

Both sides of the equation are supposed to be balanced for solving a linear equation. The equality sign denotes that the expressions on either side of the 'equal to' sign are equal. Since the equation is balanced, in order to solve it, certain mathematical operations are performed on both sides of the equation in a manner that does not affect the balance of the equation. Here is a more complex example related to a linear equation with one variable.

**Example 4**

Solve $2x-\frac{10}{2}=3(x-1)$

Step 1: Clear the fraction by dividing by 2.

$x-5=3(x-1)$

Step 2: Simplify both sides of the equation.

$x-5=3x-3$

$x=3x+2$

Step 3: Isolate x

$x-3x=2$

$-2x=2$

$x=-1$

**Example 5**

Solve $x=12(x+2)$

$x=12x+24$

Then subtract 24 on both sides of the equation.

$x-24=12x+24-24$

$x-24=12x$

Simplify.

$11x=-24$

Isolate x.

$x=-\frac{24}{11}$

**Example 6**

Solve $6x-19=3x-10$

Add 19 to both sides.

$6x-19+19=3x-10+19$

Simplify.

$6x=3x+9$

Subtract 3x from both sides.

$6x-3x=3x+9-3x$

$3x=9$

Divide both sides by 3.

$\frac{3x}{3}=\frac{9}{3}$

$x=3$

## Practice questions on solving linear equations

a. $x+5=48$

Start by subtracting 5 from each side.

$x+5-5=48-5$

Simplify by performing the subtraction.

$x=43$

b. $x-150=125$

Begin by adding 150 to each side.

$x-150+150=125+150$

Simplify by performing the addition.

$x=275$

c. Solve the equation $3x+7=19$ .

Start by subtracting 7 from each side,

$3x+7-7=19-7$

Simplify by performing the subtraction.

$3x=12$

Next, we will divide each side by 3.

$\frac{3x}{3}=\frac{12}{3}$

Finally, perform the division to reach the solution.

$x=4$

d. Solve the equation $4x+8=8x-10$

First, bring the x terms to the left side of the equation by subtracting 8x from each side.

$4x-8x-8=-10$

Then move the -8 to the right side by adding 8 to each side.

$4x-8x=-10-8$

Perform the operations on each side.

$-4x=-18$

$4x=18$

Divide each side by 4.

$\frac{4x}{4}=\frac{18}{4}$

Finally, perform the division to reach the solution.

$x=\frac{9}{2}$

e. Solve the equation $5x+5=3x+7$

Start by subtracting 5 from each side.

$5x=2-3x$

Then subtract 3x from each side.

$2x=2$

Divide each side by 2.

$\frac{2x}{2}=\frac{2}{2}$

Finally, perform the division to reach the solution.

$x=1$

f. Solve the equation $3x+4\left(x-10\right)=x+20$

First, expand the parentheses and combine like terms.

$3x+4x-40=x+20$

$7x-40=x+20$

Then add 40 to each side.

$7x=x+60$

Next, subtract x from each side.

$6x=60$

Divide both sides by 6.

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

$x=10$

g. Solve the equation $2\left(2x-5\right)=3\left(x-1\right)-4$

First, expand parentheses on both sides and combine like terms.

$4x-10=3x-7$

Add 10 to each side.

$4x=3x+3$

Subtract 3x from each side.

$4x-3x=3x+3-3x$

$x=3$

## Topics related to the Solving Linear Equations

Graphing Linear Inequalities in Two Variables

Quadratic Equations, Solving by Square Roots

Adding and Subtracting Rational Expressions with Like Denominators

## Flashcards covering the Solving Linear Equations

## Practice tests covering the Solving Linear Equations

College Algebra Diagnostic Tests

## Get help learning about solving linear equations

Working with linear equations can be difficult for many students. There are many steps and it can be tricky to keep the two sides of the equation balanced. It can also be hard to know when to add, subtract, multiply, or divide to make the sides of the equation balanced. If your student is having a hard time solving linear equations, consider having them work with a professional math or algebra tutor who can give them the 1-on-1 attention that is often lacking in the classroom.

A private tutor supports your student's in-class learning in a setting without distractions so they have an easier time focusing on each aspect of each problem. They are available to answer questions as your student comes up with them and they can supervise your student as they do their homework. This allows them to point out mistakes immediately and guide your student as they figure out the correct way to do the problems so bad habits don't take hold between class sessions. As they work together, your student's tutor can slow down and take extra time when your student needs extra help with a particular concept and can speed through concepts that your student easily understands. This makes tutoring as efficient as it is effective.

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