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Solving Linear Equations

Summary

This chapter provides comprehensive instruction on solving linear equations in one variable. Students will progress from simple one-step equations to complex multi-step equations, learning systematic approaches to isolate variables and find solutions. The chapter covers equations with variables on both sides, equations involving fractions and decimals, and literal equations where students solve for a specified variable. Students will also learn to identify different types of solutions (conditional, identity, contradiction) and apply equation-solving skills to real-world application problems.

Concepts Covered

This chapter covers the following 16 concepts from the learning graph:

  1. Solving Equations
  2. One-Step Equations
  3. Two-Step Equations
  4. Multi-Step Equations
  5. Variables on Both Sides
  6. Equations with Fractions
  7. Equations with Decimals
  8. Literal Equations
  9. Solving for a Variable
  10. Formula Manipulation
  11. Linear Equation in One Variable
  12. Solution Set
  13. Identity Equation
  14. Contradiction
  15. Conditional Equation
  16. Checking Solutions

Prerequisites

This chapter builds on concepts from:

What is a Linear Equation?

A linear equation in one variable is an equation that can be written in the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants and \(a \neq 0\). The highest power of the variable is 1, which means when graphed, these equations produce straight lines (hence "linear").

Examples of linear equations: - \(x + 5 = 12\) - \(3x - 7 = 20\) - \(\frac{2x}{3} + 4 = 10\) - \(2(x - 3) = 5x + 1\)

What makes an equation linear? - The variable appears only to the first power (no \(x^2\), \(x^3\), \(\sqrt{x}\), etc.) - The variable is not in a denominator - The variable is not inside absolute value bars, under a radical, or in an exponent

Solving an equation means finding the value(s) of the variable that make the equation true. This value is called the solution or root of the equation.

One-Step Equations

One-step equations require only a single operation to isolate the variable. They are the foundation for solving more complex equations.

Addition and Subtraction Equations

Strategy: Use the inverse operation to isolate the variable.

Example 1: Solve \(x + 7 = 15\)

Subtract 7 from both sides:

\(x + 7 - 7 = 15 - 7\)

\(x = 8\)

Check: \(8 + 7 = 15\) ✓

Example 2: Solve \(x - 9 = -3\)

Add 9 to both sides:

\(x - 9 + 9 = -3 + 9\)

\(x = 6\)

Check: \(6 - 9 = -3\) ✓

Multiplication and Division Equations

Example 3: Solve \(5x = 35\)

Divide both sides by 5:

\(\frac{5x}{5} = \frac{35}{5}\)

\(x = 7\)

Check: \(5(7) = 35\) ✓

Example 4: Solve \(\frac{x}{4} = -6\)

Multiply both sides by 4:

\(4 \cdot \frac{x}{4} = 4 \cdot (-6)\)

\(x = -24\)

Check: \(\frac{-24}{4} = -6\) ✓

Two-Step Equations

Two-step equations require two operations to solve. The standard form is \(ax + b = c\).

Strategy: 1. Use addition or subtraction to eliminate the constant term 2. Use multiplication or division to isolate the variable

Example 1: Solve \(3x + 7 = 22\)

Step 1: Subtract 7 from both sides

\(3x + 7 - 7 = 22 - 7\)

\(3x = 15\)

Step 2: Divide both sides by 3

\(\frac{3x}{3} = \frac{15}{3}\)

\(x = 5\)

Check: \(3(5) + 7 = 15 + 7 = 22\) ✓

Example 2: Solve \(\frac{x}{2} - 4 = 11\)

Step 1: Add 4 to both sides

\(\frac{x}{2} - 4 + 4 = 11 + 4\)

\(\frac{x}{2} = 15\)

Step 2: Multiply both sides by 2

\(2 \cdot \frac{x}{2} = 2 \cdot 15\)

\(x = 30\)

Check: \(\frac{30}{2} - 4 = 15 - 4 = 11\) ✓

Example 3: Solve \(-5x + 3 = 18\)

Step 1: Subtract 3 from both sides

\(-5x = 15\)

Step 2: Divide both sides by -5

\(x = -3\)

Check: \(-5(-3) + 3 = 15 + 3 = 18\) ✓

Multi-Step Equations

Multi-step equations require three or more operations to solve. These often involve the distributive property and combining like terms.

General Strategy: 1. Simplify each side (distribute and combine like terms) 2. Use addition/subtraction to collect variable terms on one side 3. Use addition/subtraction to collect constant terms on the other side 4. Use multiplication/division to isolate the variable

Example 1: Solve \(2(x + 3) - 5 = 13\)

Step 1: Distribute

\(2x + 6 - 5 = 13\)

Step 2: Combine like terms

\(2x + 1 = 13\)

Step 3: Subtract 1 from both sides

\(2x = 12\)

Step 4: Divide by 2

\(x = 6\)

Check: \(2(6 + 3) - 5 = 2(9) - 5 = 18 - 5 = 13\) ✓

Example 2: Solve \(7x - 2(x - 4) = 23\)

Step 1: Distribute the -2

\(7x - 2x + 8 = 23\)

Step 2: Combine like terms

\(5x + 8 = 23\)

Step 3: Subtract 8

\(5x = 15\)

Step 4: Divide by 5

\(x = 3\)

Check: \(7(3) - 2(3 - 4) = 21 - 2(-1) = 21 + 2 = 23\) ✓

Equations with Variables on Both Sides

When variables appear on both sides of an equation, we need to collect all variable terms on one side and all constant terms on the other.

Strategy: 1. Simplify each side if necessary 2. Move variable terms to one side (usually the side with the larger coefficient) 3. Move constant terms to the other side 4. Solve for the variable

Example 1: Solve \(5x - 7 = 2x + 8\)

Step 1: Subtract \(2x\) from both sides (to collect variables on the left)

\(5x - 2x - 7 = 2x - 2x + 8\)

\(3x - 7 = 8\)

Step 2: Add 7 to both sides

\(3x = 15\)

Step 3: Divide by 3

\(x = 5\)

Check: \(5(5) - 7 = 25 - 7 = 18\) and \(2(5) + 8 = 10 + 8 = 18\) ✓

Example 2: Solve \(3(x - 2) = 2(x + 1) + 5\)

Step 1: Distribute on both sides

\(3x - 6 = 2x + 2 + 5\)

Step 2: Combine like terms on the right

\(3x - 6 = 2x + 7\)

Step 3: Subtract \(2x\) from both sides

\(x - 6 = 7\)

Step 4: Add 6 to both sides

\(x = 13\)

Check: \(3(13 - 2) = 3(11) = 33\) and \(2(13 + 1) + 5 = 2(14) + 5 = 28 + 5 = 33\) ✓

Equations with Fractions

Equations containing fractions can be solved in two ways: work with the fractions throughout, or eliminate them by multiplying by the least common denominator (LCD).

Method 1: Multiply by the LCD (usually easier)

Example 1: Solve \(\frac{x}{3} + \frac{x}{4} = 7\)

LCD of 3 and 4 is 12. Multiply every term by 12:

\(12 \cdot \frac{x}{3} + 12 \cdot \frac{x}{4} = 12 \cdot 7\)

\(4x + 3x = 84\)

\(7x = 84\)

\(x = 12\)

Check: \(\frac{12}{3} + \frac{12}{4} = 4 + 3 = 7\) ✓

Example 2: Solve \(\frac{2x - 1}{5} = \frac{x + 2}{3}\)

LCD is 15. Multiply both sides by 15:

\(15 \cdot \frac{2x - 1}{5} = 15 \cdot \frac{x + 2}{3}\)

\(3(2x - 1) = 5(x + 2)\)

\(6x - 3 = 5x + 10\)

\(x = 13\)

Check: \(\frac{2(13) - 1}{5} = \frac{25}{5} = 5\) and \(\frac{13 + 2}{3} = \frac{15}{3} = 5\) ✓

Example 3: Solve \(\frac{3}{4}x - \frac{1}{2} = \frac{5}{6}\)

LCD is 12:

\(12 \cdot \frac{3}{4}x - 12 \cdot \frac{1}{2} = 12 \cdot \frac{5}{6}\)

\(9x - 6 = 10\)

\(9x = 16\)

\(x = \frac{16}{9}\)

Equations with Decimals

Equations with decimals can be solved by working with the decimals or by clearing them through multiplication.

Strategy to clear decimals: Multiply both sides by a power of 10 (10, 100, 1000, etc.) to eliminate all decimal points.

Example 1: Solve \(0.5x + 1.2 = 3.7\)

Multiply everything by 10 (since the longest decimal has 1 place):

\(10(0.5x) + 10(1.2) = 10(3.7)\)

\(5x + 12 = 37\)

\(5x = 25\)

\(x = 5\)

Check: \(0.5(5) + 1.2 = 2.5 + 1.2 = 3.7\) ✓

Example 2: Solve \(0.08x + 0.15(200 - x) = 0.12(200)\)

This could represent a mixture problem. Multiply by 100:

\(8x + 15(200 - x) = 12(200)\)

\(8x + 3000 - 15x = 2400\)

\(-7x + 3000 = 2400\)

\(-7x = -600\)

\(x = \frac{600}{7} \approx 85.71\)

Literal Equations and Solving for a Variable

A literal equation is an equation with two or more variables. Solving for a variable means isolating that variable on one side of the equation in terms of the other variables.

This skill is essential for formula manipulation—rearranging formulas to solve for different variables.

Example 1: Solve for \(w\) in the perimeter formula \(P = 2l + 2w\)

Goal: Isolate \(w\)

\(P = 2l + 2w\)

Subtract \(2l\) from both sides:

\(P - 2l = 2w\)

Divide by 2:

\(\frac{P - 2l}{2} = w\)

Or: \(w = \frac{P - 2l}{2}\)

Example 2: Solve for \(h\) in the area formula \(A = \frac{1}{2}bh\)

\(A = \frac{1}{2}bh\)

Multiply both sides by 2:

\(2A = bh\)

Divide by \(b\):

\(\frac{2A}{b} = h\)

Or: \(h = \frac{2A}{b}\)

Example 3: Solve for \(r\) in the simple interest formula \(I = Prt\)

\(I = Prt\)

Divide both sides by \(Pt\):

\(\frac{I}{Pt} = r\)

Or: \(r = \frac{I}{Pt}\)

Example 4: Solve for \(y\) in \(3x - 2y = 12\)

Goal: Isolate \(y\)

\(3x - 2y = 12\)

Subtract \(3x\) from both sides:

\(-2y = 12 - 3x\)

Divide by -2:

\(y = \frac{12 - 3x}{-2}\)

Simplify:

\(y = -6 + \frac{3x}{2}\)

Or: \(y = \frac{3x}{2} - 6\)

Types of Solutions

Not all linear equations have exactly one solution. There are three types:

Conditional Equations

A conditional equation is true for only certain values of the variable. Most equations we've solved are conditional.

Example: \(2x + 3 = 11\)

Solution: \(x = 4\) (true only when \(x = 4\))

The solution set is \(\{4\}\)

Identity Equations

An identity equation is true for all values of the variable. When solving, all variables cancel out and you're left with a true statement like \(5 = 5\).

Example: Solve \(3(x + 2) = 3x + 6\)

Distribute:

\(3x + 6 = 3x + 6\)

Subtract \(3x\) from both sides:

\(6 = 6\) ✓

This is always true! The solution set is all real numbers, written as \(\mathbb{R}\) or \((-\infty, \infty)\).

Contradictions

A contradiction is never true. When solving, all variables cancel out and you're left with a false statement like \(0 = 5\).

Example: Solve \(2(x + 3) = 2x + 1\)

Distribute:

\(2x + 6 = 2x + 1\)

Subtract \(2x\) from both sides:

\(6 = 1\) ✗

This is never true! There is no solution. The solution set is the empty set: \(\emptyset\) or \(\{\}\).

Summary Table

Type What Happens Solution Set Example Result
Conditional Variable isolated to one value One number \(x = 5\)
Identity Variables cancel, true statement remains All real numbers \(7 = 7\)
Contradiction Variables cancel, false statement remains No solution (empty set) \(0 = 3\)

Checking Solutions

Checking your solution is an important habit that helps catch errors and verify your answer.

To check a solution: 1. Substitute the solution back into the original equation 2. Simplify both sides 3. Verify that both sides are equal

Example: Check that \(x = 3\) is a solution to \(5x - 7 = 8\)

Substitute \(x = 3\):

\(5(3) - 7 = 8\)

\(15 - 7 = 8\)

\(8 = 8\) ✓

The solution is correct!

Why checking is important: - Catches arithmetic errors - Confirms you haven't made a sign error - Verifies that you correctly applied the distributive property - Ensures you didn't make a mistake when clearing fractions or decimals

Applications of Linear Equations

Linear equations model countless real-world situations. Setting up and solving these equations is a crucial problem-solving skill.

Number Problems

Example: The sum of three consecutive integers is 72. Find the integers.

Let \(n\) = first integer

Then \(n + 1\) = second integer

And \(n + 2\) = third integer

Equation: \(n + (n + 1) + (n + 2) = 72\)

Solve:

\(3n + 3 = 72\)

\(3n = 69\)

\(n = 23\)

The integers are 23, 24, and 25.

Check: \(23 + 24 + 25 = 72\) ✓

Geometry Problems

Example: The length of a rectangle is 5 cm more than twice its width. The perimeter is 46 cm. Find the dimensions.

Let \(w\) = width

Then \(2w + 5\) = length

Perimeter formula: \(P = 2l + 2w\)

\(46 = 2(2w + 5) + 2w\)

\(46 = 4w + 10 + 2w\)

\(46 = 6w + 10\)

\(36 = 6w\)

\(w = 6\) cm

Length = \(2(6) + 5 = 17\) cm

Check: \(P = 2(17) + 2(6) = 34 + 12 = 46\) ✓

Rate/Distance/Time Problems

Use the formula: \(\text{Distance} = \text{Rate} \times \text{Time}\)

Example: Two cars leave the same location traveling in opposite directions. One travels at 60 mph and the other at 70 mph. After how many hours will they be 390 miles apart?

Let \(t\) = time in hours

Distance apart = Distance of car 1 + Distance of car 2

\(390 = 60t + 70t\)

\(390 = 130t\)

\(t = 3\) hours

Check: Car 1: \(60(3) = 180\) miles; Car 2: \(70(3) = 210\) miles; Total: \(180 + 210 = 390\) ✓

Mixture Problems

Example: How many liters of a 30% acid solution must be mixed with 5 liters of a 60% acid solution to get a 45% acid solution?

Let \(x\) = liters of 30% solution

Amount of acid from 30% solution: \(0.30x\)

Amount of acid from 60% solution: \(0.60(5) = 3\)

Amount of acid in mixture: \(0.45(x + 5)\)

Equation: \(0.30x + 3 = 0.45(x + 5)\)

Clear decimals (multiply by 100):

\(30x + 300 = 45(x + 5)\)

\(30x + 300 = 45x + 225\)

\(75 = 15x\)

\(x = 5\) liters

Summary

In this chapter, you've developed systematic methods for solving linear equations:

Solution Strategies:

  • One-step equations: Use inverse operations
  • Two-step equations: Eliminate constant, then coefficient
  • Multi-step equations: Simplify, collect like terms, isolate variable
  • Variables on both sides: Collect variables on one side, constants on other
  • Fractions: Multiply by LCD to clear denominators
  • Decimals: Multiply by power of 10 to clear decimals

Important Concepts:

  • Literal equations: Equations with multiple variables; solve for specified variable
  • Solution types: Conditional (one solution), identity (all real numbers), contradiction (no solution)
  • Checking solutions: Always substitute back to verify

Problem-Solving Process:

  1. Read the problem carefully
  2. Define variables
  3. Write an equation based on the relationship described
  4. Solve the equation
  5. Answer the question in context
  6. Check that your answer makes sense

Linear equations are foundational to algebra and appear throughout mathematics and science. The systematic approach you've learned here will serve you well as equations become more complex in future chapters.

References

  1. One-step and two-step equations & inequalities - Khan Academy - Interactive lessons and practice problems covering solving one-step and two-step equations with all four operations, perfect for building foundational equation-solving skills.