Solving Multistep Linear Equations with Variables on Both Sides

Hello, and welcome back. In the previous lesson, you evaluated expressions by following the order of operations and substituting values carefully. Those habits now become tools for solving equations: we will simplify expressions accurately, but this time we will find the value that makes two expressions equal.

By the end of this lesson, you should be able to solve multistep linear equations even when the variable appears on both sides of the equals sign, and verify that your answer works. Plan on about 40 minutes, including two short videos.

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Equality is a balance

An equation such as

$$5x-4=2x+11$$

claims that the expression on the left and the expression on the right have the same value. A solution is a value of $x$ that makes that claim true.

Think of the equals sign as a balanced scale. You may add, subtract, multiply, or divide, but you must perform the same operation on both sides. If you change only one side, the equation is no longer guaranteed to remain true.

The new challenge is that $x$ appears on both sides. The goal is not to “move” a term across the equal sign by magic. Instead, use an inverse operation on both sides so that the variable terms end up on one side and the constant terms on the other.

The coefficient is the number multiplying a variable. For example, the coefficient of $7x$ is $7$, and the coefficient of $-3x$ is $-3$.

Consider:

$$7x+3=4x+18$$

The variable terms are $7x$ and $4x$. We can eliminate $4x$ from the right by subtracting $4x$ from both sides:

$$7x+3-4x=4x+18-4x$$

Combine like terms:

$$3x+3=18$$

Now remove the $+3$ by subtracting $3$ from both sides:

$$3x+3-3=18-3$$ $$3x=15$$

Finally, divide both sides by $3$:

$$\frac{3x}{3}=\frac{15}{3}$$ $$\boxed{x=5}$$

The choice to subtract $4x$, rather than $7x$, was strategic: it left a positive $3x$. Subtracting $7x$ would also be valid, but it would produce $-3x$, which adds opportunities for sign errors.

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A reliable solving strategy

For a multistep linear equation with variables on both sides, use this order:

1. Simplify each side first. Distribute through parentheses and combine like terms where possible.
2. Choose a variable side. Eliminate the variable term from the other side using addition or subtraction on both sides.
3. Gather constants on the opposite side. Use addition or subtraction on both sides.
4. Make the coefficient $1$. Divide or multiply both sides as needed.
5. Check the answer in the original equation.

The word “first” matters. Do not try to collect variables before distributing or combining like terms when those simplifications are available. The previous lesson’s order-of-operations skills are doing important work here.

Situation	Appropriate action
$+6x$ appears on a side and you want to remove it	Subtract $6x$ from both sides
$-6x$ appears on a side and you want to remove it	Add $6x$ to both sides
$+9$ appears with the variable	Subtract $9$ from both sides
$-9$ appears with the variable	Add $9$ to both sides
You have $5x=35$	Divide both sides by $5$

A common shorthand says that a term “crosses the equals sign and changes sign.” This may help you remember the arithmetic, but it hides the reason. What actually happens is an inverse operation on both sides. Writing that operation explicitly will make your work more dependable.

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Equations that need simplifying first

Parentheses and like terms make an equation multistep. Simplify them before collecting variable terms.

Now work through a similar equation:

$$4(x-2)+3x=5x+10$$

1. Distribute

The $4$ multiplies both terms inside the parentheses:

$$4x-8+3x=5x+10$$

A frequent error is writing $4(x-2)=4x-2$. The $-2$ must also be multiplied by $4$, producing $-8$.

2. Combine like terms

On the left, $4x$ and $3x$ are like terms:

$$7x-8=5x+10$$

3. Put all variable terms on one side

Subtract $5x$ from both sides:

$$7x-8-5x=5x+10-5x$$ $$2x-8=10$$

4. Put constants on the other side

Add $8$ to both sides:

$$2x-8+8=10+8$$ $$2x=18$$

Divide both sides by $2$:

$$\frac{2x}{2}=\frac{18}{2}$$ $$\boxed{x=9}$$

5. Check in the original equation

Use $x=9$, not the simplified equation, for the most reliable check:

$$4(9-2)+3(9)=5(9)+10$$

Evaluate the left side:

$$4(7)+27=28+27=55$$

Evaluate the right side:

$$45+10=55$$

Both sides equal $55$, so $x=9$ is correct.

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When there is not one numerical solution

Most equations in this lesson have exactly one solution. But after simplifying, some linear equations reveal that they have no solution or infinitely many solutions.

No solution

Consider:

$$2(x+1)+x=3x+7$$

Distribute and combine like terms:

$$2x+2+x=3x+7$$ $$3x+2=3x+7$$

Subtract $3x$ from both sides:

$$2=7$$

This statement is false. No value of $x$ can make it true, so the equation has no solution.

Infinitely many solutions

Now consider:

$$3(x+2)-x=2x+6$$

Distribute and combine:

$$3x+6-x=2x+6$$ $$2x+6=2x+6$$

Subtract $2x$ from both sides:

$$6=6$$

This statement is always true. Every real number makes the original equation true, so the equation has infinitely many solutions.

These outcomes are not mistakes when they appear at the end of valid algebra. They tell you something real about the equation:

Final statement	Meaning
$x=5$	One solution
$2=7$	No solution
$6=6$	Infinitely many solutions; all real numbers work

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Error checks that save time

Before finalizing a solution, look for these common issues:

• Changing only one side: Every operation used to preserve equality must be performed on both sides.
• Incorrect distribution: Multiply the outside factor by every term inside the parentheses, including negative terms.
• Combining unlike terms: $3x+5$ cannot become $8x$. The terms have different variable parts.
• Losing a negative sign: Keep subtraction signs attached to the terms that follow them.
• Stopping too early: $4x=28$ is not yet solved; divide by $4$.
• Checking a simplified version only: Substitute into the original equation when possible.

A clean layout helps: write one equivalent equation per line, show the operation applied to both sides, and simplify before beginning the next operation.

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Key takeaways

To solve multistep linear equations with variables on both sides:

1. Simplify each side by distributing and combining like terms.
2. Use the same addition or subtraction operation on both sides to collect variable terms on one side.
3. Use the same operation on both sides to move constants to the other side.
4. Divide by the variable’s coefficient.
5. Check by substituting the answer into the original equation.

Choosing to eliminate the variable term with the smaller coefficient is often convenient, but it is a strategy rather than a rule. The nonnegotiable rule is that every operation must preserve equality by being performed on both sides.

Next, you will extend this balance-based reasoning to linear inequalities. The process begins similarly, but multiplying or dividing by a negative number introduces one important new rule.