← MATH 114 MathScape 0 MATH114

Solving Linear Equations in One Variable

8 min read

Jump to a section

Textbook Reference

Primary source OpenStax College Algebra 2e, Section 2.2: “Linear Equations in One Variable”
Direct link https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable

This source is free and openly licensed. College Algebra 2e Section 2.2 covers linear equations, including those with fractions.


Key idea

An equation is a balance scale, and solving it is the act of stripping away everything that surrounds the variable while keeping the scale balanced.

The two sides of an equation weigh the same. Whatever you do to one side you must do to the other, or the balance breaks. To find the value of $x$, you peel off the operations wrapped around it one at a time, always undoing the outermost operation first. This is the order of operations run in reverse: you built the expression by following PEMDAS, so you unbuild it by going the opposite direction.

This means: solving is unwrapping. Identify what is being done to $x$, then undo it in reverse order, doing the same thing to both sides at every step. The goal is to leave $x$ standing alone.


Prerequisite Check

Before this lesson, make sure you can do all of the following:

If simplifying is shaky, review Simplifying Algebraic Expressions first.


Quick Reference

A linear equation in one variable can be written in the form \[ ax + b = 0 \quad \text{where } a \neq 0. \] Its solution is the single value of $x$ that makes the equation true.

The properties of equality (the legal moves). For any real numbers, if $A = B$ then:

Move Statement
Add the same amount $A + c = B + c$
Subtract the same amount $A - c = B - c$
Multiply by the same nonzero amount $A \cdot c = B \cdot c$
Divide by the same nonzero amount $\dfrac{A}{c} = \dfrac{B}{c}$

The solving procedure (flashcard).

  1. Clear fractions by multiplying every term by the common denominator (optional but tidy).
  2. Distribute to remove parentheses.
  3. Combine like terms on each side.
  4. Move all variable terms to one side, all constants to the other (add or subtract).
  5. Divide by the coefficient of the variable.
  6. Check by substituting back into the original equation.

Key Concepts

1. What an Equation Is

An equation states that two expressions are equal. A solution is a value of the variable that makes the statement true. Solving means finding every such value.

Example 1. Verify that $x = 4$ is a solution of $2x + 1 = 9$.

Substitute: $2(4) + 1 = 8 + 1 = 9$. The left side equals the right side, so $x = 4$ is a solution.

A linear equation in one variable has exactly one solution (unless it is a special case, covered in Section 5).


2. Undoing One Operation at a Time

To isolate the variable, undo the operations wrapped around it, doing the same thing to both sides. Undo the outermost operation first.

Example 2. Solve $x + 7 = 12$.

The variable has $7$ added to it. Undo by subtracting $7$ from both sides: \[ x + 7 - 7 = 12 - 7 \implies x = 5 \]

Example 3. Solve $3x = 21$.

The variable is multiplied by $3$. Undo by dividing both sides by $3$: \[ \frac{3x}{3} = \frac{21}{3} \implies x = 7 \]

Important: whatever you do to one side, do to the other. Subtracting $7$ from only the left side breaks the balance and gives a wrong answer. The equality is preserved only when both sides change identically.


3. Two-Step Equations

When more than one operation surrounds the variable, undo them in reverse order: undo addition or subtraction before multiplication or division.

Example 4. Solve $2x + 5 = 17$.

Step 1. Undo the $+5$: subtract $5$ from both sides. $2x = 12$.

Step 2. Undo the $\times 2$: divide both sides by $2$. $x = 6$.

Check: $2(6) + 5 = 12 + 5 = 17$. Correct.

Important: undo addition before division. A frequent error is dividing $2x + 5 = 17$ by $2$ first, writing $x + 5 = 8.5$. The $+5$ is not inside the multiplication, so it must be removed first. Reverse PEMDAS: subtraction before division.


4. Variables on Both Sides

When the variable appears on both sides, first move all variable terms to one side and all constants to the other.

Example 5. Solve $5x - 3 = 2x + 9$.

Step 1. Subtract $2x$ from both sides to gather variables on the left: $3x - 3 = 9$.

Step 2. Add $3$ to both sides: $3x = 12$.

Step 3. Divide by $3$: $x = 4$.

Check: Left $5(4) - 3 = 17$. Right $2(4) + 9 = 17$. Correct.


5. Equations with Parentheses and Fractions

Distribute to clear parentheses, and multiply through by a common denominator to clear fractions, before solving.

Example 6. Solve $3(x - 2) = 2x + 4$.

Distribute: $3x - 6 = 2x + 4$. Subtract $2x$: $x - 6 = 4$. Add $6$: $x = 10$.

Check: $3(10 - 2) = 3(8) = 24$ and $2(10) + 4 = 24$. Correct.

Example 7. Solve $\dfrac{x}{2} + \dfrac{1}{3} = \dfrac{5}{6}$.

The common denominator is $6$. Multiply every term by $6$: \[ 6 \cdot \frac{x}{2} + 6 \cdot \frac{1}{3} = 6 \cdot \frac{5}{6} \implies 3x + 2 = 5 \] Subtract $2$: $3x = 3$. Divide by $3$: $x = 1$.

Check: $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}$. Correct.


6. Special Cases: No Solution and All Reals

Not every linear equation has exactly one solution. After simplifying, two special outcomes can appear.

Example 8. Solve $2(x + 3) = 2x + 6$.

Distribute: $2x + 6 = 2x + 6$. Subtract $2x$: $6 = 6$. This is always true, so the equation is an identity and every real number is a solution.

Example 9. Solve $x + 4 = x + 9$.

Subtract $x$: $4 = 9$. This is false, so the equation has no solution.

Self-check: Solve $4(x - 1) = 2(x + 3)$.

Distribute: $4x - 4 = 2x + 6$. Subtract $2x$: $2x - 4 = 6$. Add $4$: $2x = 10$. Divide by $2$: $x = 5$. Check: $4(5 - 1) = 16$ and $2(5 + 3) = 16$. Correct.


Common Errors Summary

Error Example Correction
Changing only one side $x + 7 = 12 \to x = 12$ Subtract $7$ from BOTH sides: $x = 5$
Dividing before subtracting $2x + 5 = 17 \to x + 5 = 8.5$ Subtract $5$ first: $2x = 12$, then $x = 6$
Forgetting to distribute $3(x - 2) = 3x - 2$ Distribute the $3$: $3x - 6$
Clearing only some fraction terms multiplying two of three terms by 6 Multiply EVERY term by the common denominator
Misreading an identity as no solution $6 = 6$ called “no solution” A true statement means all reals are solutions

Common Misconceptions

Common misconception

an operation applied to one side of an equation does not need to be applied to the other side.

This is the multiplicative-not-additive error in the sense of balance. An equation is a statement of equality; performing an operation on only one side destroys that equality. For example, solving $2x + 5 = 17$ by dividing only the left side by $2$ produces $x + 5 = 17$, which is incorrect. Every inverse operation must be applied identically to both sides so that the balance is preserved at every step.


Leveled Practice

Level 1 -- Direct Application

Problem 1. Solve $x - 8 = 3$.

Show answer

Add $8$ to both sides: $x = 11$.


Problem 2. Solve $4x = 28$.

Show answer

Divide both sides by $4$: $x = 7$.


Problem 3. Solve $3x - 5 = 16$.

Show answer

Add $5$: $3x = 21$. Divide by $3$: $x = 7$. Check: $3(7) - 5 = 16$. Correct.


Level 2 -- Multiple Steps

Problem 4. Solve $7x + 2 = 4x + 17$.

Show answer

Subtract $4x$: $3x + 2 = 17$. Subtract $2$: $3x = 15$. Divide by $3$: $x = 5$.


Problem 5. Solve $2(3x - 4) = 4x + 6$.

Show answer

Distribute: $6x - 8 = 4x + 6$. Subtract $4x$: $2x - 8 = 6$. Add $8$: $2x = 14$. Divide: $x = 7$.


Level 3 -- Fractions and Special Cases

Problem 6. Solve $\dfrac{x}{3} - \dfrac{1}{2} = \dfrac{x}{6}$.

Show answer

Common denominator $6$. Multiply every term: $2x - 3 = x$. Subtract $x$: $x - 3 = 0$. Add $3$: $x = 3$. Check: $\dfrac{3}{3} - \dfrac{1}{2} = 1 - 0.5 = 0.5$ and $\dfrac{3}{6} = 0.5$. Correct.


Problem 7. Solve $3(x + 2) = 3x + 5$ and interpret the result.

Show answer

Distribute: $3x + 6 = 3x + 5$. Subtract $3x$: $6 = 5$. This is false, so the equation has no solution. No value of $x$ can make $6$ equal $5$.


Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:


Mental Model

Think of an equation as a balance scale and the variable as a gift buried inside layers of wrapping paper.

The two pans of the scale weigh the same. To find the gift you remove wrapping one layer at a time, and to keep the scale level you must remove the same weight from both pans at every step. You unwrap in the reverse of the order the gift was wrapped: the order of operations built the expression, so you peel it off in reverse, additions and subtractions before multiplications and divisions. When the last layer comes off, the variable stands alone and its value is revealed. If the variable disappears entirely, the scale is telling you something special: a true leftover means every number balances (an identity), and a false leftover means nothing balances (no solution).


Connections

Within MATH 114

Toward Later Mathematics

Audience Notes

For students who find math intimidating: There is one idea to hold onto: do the same thing to both sides, and undo the outer operation first. Every problem is the same dance with different numbers. Checking your answer at the end turns “I hope this is right” into “I know this is right”.

For students interested in proof: Each step is justified by a named property of equality. Writing the property beside each line turns a calculation into a short proof that the final value really is the solution.

For students interested in careers: Solving for an unknown is the most common quantitative task in any technical field: rearranging a formula to isolate the quantity you need is exactly this skill, applied to letters instead of a lone $x$.


Back to MATH 114 Skills | Previous: Simplifying Algebraic Expressions | Next: The Rectangular Coordinate Plane