Solving Linear Equations

Learn to solve one-step, two-step, and multi-step linear equations

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Solving Linear Equations

What Is a Linear Equation?

A linear equation is an equation in which the variable appears only to the first power โ€” no exponents, no square roots, no variables multiplied together. The word "linear" comes from the fact that the graph of every such equation is a straight line.

The standard form of a linear equation in one variable looks like ax+b=cax + b = c, where aa, bb, and cc are real numbers and aโ‰ 0a \neq 0. The letter xx is the unknown variable โ€” the value we are trying to find.

Some examples of linear equations:

  • 2x+5=132x + 5 = 13
  • 3xโˆ’7=203x - 7 = 20
  • x4+3=7\frac{x}{4} + 3 = 7
  • 5(xโˆ’2)=155(x - 2) = 15

And here are some NON-examples (these are NOT linear):

  • x2+3=12x^2 + 3 = 12 โ€” the variable is squared, making it a quadratic equation
  • x=5\sqrt{x} = 5 โ€” the variable is under a radical
  • xy=10xy = 10 โ€” two variables multiplied together

Being able to tell the difference between linear and nonlinear equations is the first step toward mastering algebra.

The Golden Rule: Keep the Equation Balanced

Think of an equation as a balance scale with the equals sign as the fulcrum. Whatever sits on the left side must weigh the same as the right side. This is the most important idea in all of equation-solving:

Whatever you do to one side, you must do to the other side.

This principle is captured by four properties of equality:

| Property | Statement | What it means | |----------|-----------|---------------| | Addition Property | If a=ba = b, then a+c=b+ca + c = b + c | Add the same value to both sides | | Subtraction Property | If a=ba = b, then aโˆ’c=bโˆ’ca - c = b - c | Subtract the same value from both sides | | Multiplication Property | If a=ba = b, then ac=bcac = bc | Multiply both sides by the same value | | Division Property | If a=ba = b and cโ‰ 0c \neq 0, then ac=bc\frac{a}{c} = \frac{b}{c} | Divide both sides by the same nonzero value |

All four properties keep the equation balanced and equivalent to the original. Any sequence of these operations will never add or remove a solution.

Inverse Operations: Your Equation-Solving Toolkit

The ultimate goal of solving an equation is to isolate the variable โ€” get the variable completely alone on one side of the equation with a coefficient of 1. To peel numbers away from the variable, you use inverse operations โ€” operations that undo each other:

  • Addition and subtraction are inverses of each other
  • Multiplication and division are inverses of each other

Think of it like peeling layers off an onion: each inverse operation removes one layer, bringing you closer to xx alone. The order matters โ€” you undo operations in the reverse order of operations (reverse PEMDAS), meaning you undo addition and subtraction before multiplication and division.

Solving One-Step Equations

A one-step equation requires exactly one operation to solve. These are the building blocks for every harder equation you will encounter, so mastering them is critical.

Undo Addition โ€” subtract from both sides: x+7=12x + 7 = 12 x+7โˆ’7=12โˆ’7x + 7 - 7 = 12 - 7 x=5x = 5

Undo Subtraction โ€” add to both sides: xโˆ’4=9x - 4 = 9 xโˆ’4+4=9+4x - 4 + 4 = 9 + 4 x=13x = 13

Undo Multiplication โ€” divide both sides: 3x=153x = 15 3x3=153\frac{3x}{3} = \frac{15}{3} x=5x = 5

Undo Division โ€” multiply both sides: x2=8\frac{x}{2} = 8 2โ‹…x2=2โ‹…82 \cdot \frac{x}{2} = 2 \cdot 8 x=16x = 16

Always verify by plugging back into the original: 162=8\frac{16}{2} = 8 โœ“

Solving Two-Step Equations

A two-step equation has two operations applied to the variable. The strategy is to undo addition or subtraction first, then undo multiplication or division โ€” this is the reverse PEMDAS approach.

Example 1: 2x+5=132x + 5 = 13

Step 1 โ€” Undo the +5+5 by subtracting from both sides: 2x=82x = 8

Step 2 โ€” Undo the ร—2\times 2 by dividing both sides: x=4x = 4

Check: 2(4)+5=8+5=132(4) + 5 = 8 + 5 = 13 โœ“

Example 2: 3xโˆ’7=203x - 7 = 20

Step 1 โ€” Add 7: 3x=273x = 27

Step 2 โ€” Divide by 3: x=9x = 9

Check: 3(9)โˆ’7=27โˆ’7=203(9) - 7 = 27 - 7 = 20 โœ“

Example 3: x4+3=7\frac{x}{4} + 3 = 7

Step 1 โ€” Subtract 3: x4=4\frac{x}{4} = 4

Step 2 โ€” Multiply by 4: x=16x = 16

Check: 164+3=4+3=7\frac{16}{4} + 3 = 4 + 3 = 7 โœ“

The two-step pattern appears constantly in algebra, so practice until it feels automatic.

The Distributive Property in Equations

Many multi-step equations involve parentheses. Before anything else, you must distribute the outside factor โ€” multiply it by every term inside the parentheses:

a(b+c)=ab+aca(b + c) = ab + ac

Example: Solve 5(xโˆ’2)=155(x - 2) = 15

Step 1 โ€” Distribute the 5: 5xโˆ’10=155x - 10 = 15

Step 2 โ€” Add 10: 5x=255x = 25

Step 3 โ€” Divide by 5: x=5x = 5

Common mistake alert! Always distribute to ALL terms inside. For instance, 3(x+2)3(x + 2) equals 3x+63x + 6, NOT 3x+23x + 2. Forgetting to distribute to the second term is one of the most frequent errors in algebra and is called an incomplete distribution error.

Combining Like Terms

Before isolating the variable, simplify each side of the equation by combining like terms. Like terms have the same variable raised to the same power โ€” for example, 4x4x and โˆ’x-x are like terms, but 3x3x and 3x23x^2 are not.

Example: Solve 4x+3โˆ’x+7=224x + 3 - x + 7 = 22

Combine like terms on the left: 4xโˆ’x=3x4x - x = 3x and 3+7=103 + 7 = 10 3x+10=223x + 10 = 22

Subtract 10: 3x=123x = 12

Divide by 3: x=4x = 4

Simplifying before solving always makes the work easier. A good habit is to scan both sides of the equation for like terms before doing anything else, which is called simplifying each side first.

Variables on Both Sides

When the variable appears on both sides of the equation, your first job is to collect all variable terms on one side and all constant terms on the other. It does not matter which side you choose โ€” pick whichever makes the arithmetic easier.

Example 1: 5xโˆ’3=2x+125x - 3 = 2x + 12

Step 1 โ€” Subtract 2x2x from both sides to get variables on the left: 3xโˆ’3=123x - 3 = 12

Step 2 โ€” Add 3: 3x=153x = 15

Step 3 โ€” Divide by 3: x=5x = 5

Check: Left = 5(5)โˆ’3=225(5) - 3 = 22. Right = 2(5)+12=222(5) + 12 = 22 โœ“

Example 2: 2(x+3)โˆ’5=4xโˆ’72(x + 3) - 5 = 4x - 7

Distribute: 2x+6โˆ’5=4xโˆ’72x + 6 - 5 = 4x - 7, which simplifies to 2x+1=4xโˆ’72x + 1 = 4x - 7

Subtract 2x2x: 1=2xโˆ’71 = 2x - 7

Add 7: 8=2x8 = 2x

Divide by 2: x=4x = 4

The golden strategy is to move variables to one side and constants to the other using inverse operations.

Clearing Fractions with the LCD

Equations with fractions can look intimidating, but there is a powerful shortcut: multiply every term on both sides by the least common denominator to eliminate all fractions at once. This technique is called clearing the denominators and turns a messy problem into a clean one.

Example 1: x3+x4=7\frac{x}{3} + \frac{x}{4} = 7

The LCD of 3 and 4 is 12. Multiply every term by 12: 12โ‹…x3+12โ‹…x4=12โ‹…712 \cdot \frac{x}{3} + 12 \cdot \frac{x}{4} = 12 \cdot 7 4x+3x=844x + 3x = 84 7x=847x = 84 x=12x = 12

Example 2: 2x+13=5\frac{2x + 1}{3} = 5

Multiply both sides by 3: 2x+1=152x + 1 = 15 2x=142x = 14 x=7x = 7

You can also clear decimal coefficients by multiplying both sides by a power of 10. For example, 0.3x+1.5=4.20.3x + 1.5 = 4.2 becomes 3x+15=423x + 15 = 42 when you multiply everything by 10.

Special Cases: No Solution and Infinitely Many Solutions

Most linear equations have exactly one unique solution. However, two special cases can occur:

Identity โ€” Infinitely Many Solutions: An identity equation is true for every possible value of the variable. Example: 2(x+3)=2x+62(x + 3) = 2x + 6 2x+6=2x+62x + 6 = 2x + 6 6=6(alwaysย true)6 = 6 \quad \text{(always true)}

When the variables cancel and you get a true statement like 0=00 = 0, every real number is a solution.

Contradiction โ€” No Solution: A contradiction equation is never true for any value. Example: 2(x+3)=2x+82(x + 3) = 2x + 8 2x+6=2x+82x + 6 = 2x + 8 6=8(false!)6 = 8 \quad \text{(false!)}

When the variables cancel and you get a false statement like 6=86 = 8, no value of xx can make the equation true. The equation has no solution.

Recognizing these special cases saves time โ€” once the variable terms cancel, the answer is immediate.

Common Mistakes and How to Avoid Them

Here are the errors that trip up students most often. Awareness is the first step to avoiding them!

  1. Performing operations on only one side โ€” If you add 5 to the left, you MUST add 5 to the right. The equation stays balanced only when you apply the same operation to both sides.

  2. Distributing to only the first term โ€” 3(x+2)=3x+63(x + 2) = 3x + 6, NOT 3x+23x + 2. Multiply the factor by every single term inside the parentheses.

  3. Sign errors when subtracting โ€” Remember that subtracting a negative is the same as adding: aโˆ’(โˆ’b)=a+ba - (-b) = a + b. Always double-check sign changes.

  4. Confusing subtraction with division โ€” In 2x=102x = 10, divide by 2 to get x=5x = 5. Do NOT subtract 2 to get x=8x = 8. The operation between 2 and xx is multiplication, so undo with division.

  5. Skipping verification โ€” Always substitute your answer back into the original equation. This catches arithmetic mistakes and reveals special cases. The habit of checking your solution is what separates strong algebra students from the rest.

Word Problems: Translating English into Algebra

Linear equations model countless real-world situations. The key skill is translating words into equations by identifying the unknown and writing a mathematical statement.

Age Problem: "Maya is 5 years older than Noah. Their ages sum to 27. How old is Noah?" Let xx = Noah's age. Maya's age is x+5x + 5. x+(x+5)=27x + (x + 5) = 27 2x+5=27โ€…โ€ŠโŸนโ€…โ€Š2x=22โ€…โ€ŠโŸนโ€…โ€Šx=112x + 5 = 27 \implies 2x = 22 \implies x = 11 Noah is 11, Maya is 16.

Money Problem: "You have 50 dollars. After buying notebooks at 7 dollars each, you have 15 dollars left. How many did you buy?" 50โˆ’7x=15โ€…โ€ŠโŸนโ€…โ€Š7x=35โ€…โ€ŠโŸนโ€…โ€Šx=550 - 7x = 15 \implies 7x = 35 \implies x = 5 You bought 5 notebooks.

Geometry Problem: "The perimeter of a rectangle is 40 cm. The length is 3 cm more than the width." Let ww = width. Length =w+3= w + 3. 2w+2(w+3)=40โ€…โ€ŠโŸนโ€…โ€Š4w+6=40โ€…โ€ŠโŸนโ€…โ€Š4w=34โ€…โ€ŠโŸนโ€…โ€Šw=8.52w + 2(w + 3) = 40 \implies 4w + 6 = 40 \implies 4w = 34 \implies w = 8.5 Width = 8.5 cm, Length = 11.5 cm.

Distance-Rate-Time: Two cars leave the same point in opposite directions at 55 mph and 65 mph. When are they 360 miles apart? 55t+65t=360โ€…โ€ŠโŸนโ€…โ€Š120t=360โ€…โ€ŠโŸนโ€…โ€Št=3ย hours55t + 65t = 360 \implies 120t = 360 \implies t = 3 \text{ hours}

The problem-solving strategy is always the same: define a variable, write an equation, solve, then check that the answer makes sense in context.

Step-by-Step Solving Strategy

Follow this systematic approach whenever you face any linear equation:

  1. Simplify each side โ€” distribute and combine like terms
  2. Collect variable terms on one side using addition or subtraction
  3. Collect constant terms on the other side
  4. Divide or multiply to make the coefficient of xx equal to 1
  5. Check your answer by substituting back into the original equation

This five-step framework works for every linear equation, from the simplest one-step problem to the most complex multi-step challenge.

Quick Reference Table

| Equation Type | Example | First Move | |--------------|---------|-----------| | One-step (addition) | x+5=12x + 5 = 12 | Subtract 5 from both sides | | One-step (multiplication) | 3x=153x = 15 | Divide both sides by 3 | | Two-step | 2x+5=132x + 5 = 13 | Subtract 5, then divide by 2 | | Distribution needed | 3(xโˆ’2)=153(x - 2) = 15 | Distribute the 3 first | | Variables on both sides | 5xโˆ’3=2x+125x - 3 = 2x + 12 | Move variable terms to one side | | Fractions | x3+x4=7\frac{x}{3} + \frac{x}{4} = 7 | Multiply all terms by the LCD | | Decimals | 0.3x+1.5=4.20.3x + 1.5 = 4.2 | Multiply all terms by 10 |

Master each row of this table and you will be ready for any linear equation that comes your way!

๐Ÿ“š Practice Problems

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