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4.3 Solving Linear Equations & Inequalities

Key Takeaways

  • The balance model dictates that whatever mathematical operation is applied to one side of an equation or inequality must be identically applied to the opposite side.

  • Solving linear equations involves isolating the variable by applying inverse operations in reverse order of operations (reverse BEDMAS).

  • Equations containing fractional denominators are simplified by multiplying every term on both sides by the Least Common Denominator (LCD).

  • When solving linear inequalities, multiplying or dividing both sides by a negative number flips the direction of the inequality sign (< becomes >, and vice versa).

  • Verification of algebraic solutions requires substituting the calculated value back into the original equation to confirm that Left-Hand Side (LHS) equals Right-Hand Side (RHS).

Last updated: August 2026

4.3 Solving Linear Equations & Inequalities

Conceptual Foundation: The Balance Model & Inverse Operations

Solving linear equations and inequalities is a core expectation of Ontario middle-school and secondary mathematics curriculum. On the MPT, candidate teachers must demonstrate procedural accuracy, conceptual understanding of equation balance, and mastery of algebraic properties.

An equation is a mathematical statement asserting that two expressions are equal in value (LHS=RHSLHS = RHS). The balance scale model provides the foundational intuition: an equation is like a balanced two-pan scale. To maintain balance, any mathematical operation performed on the left-hand side (LHS) must be identically performed on the right-hand side (RHS).

Inverse Operations

To isolate a variable, we perform inverse operations in reverse order of operations (reverse BEDMAS):

  • Addition (+\mathbf{+}) is inverted by Subtraction (−\mathbf{-})
  • Subtraction (−\mathbf{-}) is inverted by Addition (+\mathbf{+})
  • Multiplication (×\mathbf{\times}) is inverted by Division (÷\mathbf{\div})
  • Division (÷\mathbf{\div}) is inverted by Multiplication (×\mathbf{\times})

Solving 1-Step and 2-Step Linear Equations

1-Step Equations

A single inverse operation isolates the variable.

  • Example: x+14=31  ⟹  x=31−14  ⟹  x=17x + 14 = 31 \implies x = 31 - 14 \implies x = 17
  • Example: −6x=42  ⟹  x=42−6  ⟹  x=−7-6x = 42 \implies x = \frac{42}{-6} \implies x = -7

2-Step Equations

Requires undoing addition/subtraction first, followed by multiplication/division.

Solve: 4x−9=27\text{Solve: } 4x - 9 = 27
  1. Add 9 to both sides: 4x=27+9  ⟹  4x=364x = 27 + 9 \implies 4x = 36
  2. Divide both sides by 4: x=364  ⟹  x=9x = \frac{36}{4} \implies x = 9

Multi-Step Equations with Variables on Both Sides

When variables appear on both sides of the equals sign, use the following systematic 5-step algorithm:

Systematic Solving Algorithm:

  1. Clear Parentheses: Apply the distributive property to expand all brackets.
  2. Simplify Each Side: Combine like terms independently on the LHS and RHS.
  3. Collect Variable Terms: Add or subtract variable terms to move all variable terms to one side of the equation.
  4. Collect Constant Terms: Add or subtract constants to move all numerical terms to the opposite side.
  5. Isolate the Variable: Divide or multiply by the variable's coefficient to solve for xx.

Example Worked Problem:

Solve: 5(x−2)+4=2(2x+7)−3\text{Solve: } 5(x - 2) + 4 = 2(2x + 7) - 3
  • Step 1 (Expand): 5x−10+4=4x+14−35x - 10 + 4 = 4x + 14 - 3
  • Step 2 (Combine like terms): 5x−6=4x+115x - 6 = 4x + 11
  • Step 3 (Collect variables on LHS): Subtract 4x4x from both sides: 5x−4x−6=11  ⟹  x−6=115x - 4x - 6 = 11 \implies x - 6 = 11
  • Step 4 (Collect constants on RHS): Add 6 to both sides: x=11+6  ⟹  x=17x = 11 + 6 \implies x = 17
  • Step 5 (Verification): Substitute x=17x = 17 into original equation:
    • LHS=5(17−2)+4=5(15)+4=75+4=79\text{LHS} = 5(17 - 2) + 4 = 5(15) + 4 = 75 + 4 = 79
    • RHS=2(2(17)+7)−3=2(34+7)−3=2(41)−3=82−3=79\text{RHS} = 2(2(17) + 7) - 3 = 2(34 + 7) - 3 = 2(41) - 3 = 82 - 3 = 79
    • Since LHS=RHS=79\text{LHS} = \text{RHS} = 79, the solution x=17x = 17 is verified.

Clearing Fractional Denominators Using LCD

Equations containing fractions are a frequent source of error on the MPT. The most efficient strategy is to clear all fractions in Step 1 by multiplying every term on both sides of the equation by the Least Common Denominator (LCD) of all denominators present.

Step-by-Step LCD Fraction Clearing Technique:

Given:

2x+13−x−24=2\frac{2x + 1}{3} - \frac{x - 2}{4} = 2
  1. Identify the LCD: Denominators are 3 and 4. The LCD is 12.
  2. Multiply EVERY term by 12: 12⋅(2x+13)−12⋅(x−24)=12⋅212 \cdot \left(\frac{2x + 1}{3}\right) - 12 \cdot \left(\frac{x - 2}{4}\right) = 12 \cdot 2
  3. Simplify numerator multipliers: 4(2x+1)−3(x−2)=244(2x + 1) - 3(x - 2) = 24
  4. Expand brackets: 8x+4−3x+6=248x + 4 - 3x + 6 = 24
  5. Combine like terms: 5x+10=245x + 10 = 24
  6. Isolate variable: 5x=24−10  ⟹  5x=14  ⟹  x=145=2.85x = 24 - 10 \implies 5x = 14 \implies x = \frac{14}{5} = 2.8

Solving Linear Inequalities & The Negative Division Rule

A linear inequality uses order symbols (<,≤,>,≥<, \le, >, \ge) instead of an equals sign. The solution set represents a range of values rather than a single number.

Rules of Inequality Manipulation

All equation balance operations apply to inequalities with one vital exception:

The Golden Rule of Inequalities: When you multiply or divide both sides of an inequality by a negative number, you MUST reverse (flip) the direction of the inequality sign.

Mathematical Proof of Sign Flipping:

Consider the true statement: 2<52 < 5.

  • If we multiply both sides by −1-1:
    • Left side becomes −2-2. Right side becomes −5-5.
    • On a number line, −2-2 is to the right of −5-5, so −2>−5-2 > -5.
    • Maintaining 2<5  ⟹  −2<−52 < 5 \implies -2 < -5 would be mathematically false! Thus, the sign must reverse.
Inequality SymbolMeaningExample SolutionGraphic Representation on Number Line
<<Less thanx<4x < 4Open circle at 4, arrow pointing left
≤\leLess than or equal tox≤4x \le 4Closed/solid circle at 4, arrow pointing left
>>Greater thanx>4x > 4Open circle at 4, arrow pointing right
≥\geGreater than or equal tox≥4x \ge 4Closed/solid circle at 4, arrow pointing right

Step-by-Step Worked Equation & Inequality Solutions

Problem 1: Equation with Decimals and Distribution

Problem: Solve for xx: 0.4(2x−5)=1.2x−3.60.4(2x - 5) = 1.2x - 3.6

Step-by-Step Solution:

  • Option A: Clear decimals by multiplying entire equation by 10. 10⋅[0.4(2x−5)]=10⋅(1.2x−3.6)10 \cdot [0.4(2x - 5)] = 10 \cdot (1.2x - 3.6) 4(2x−5)=12x−364(2x - 5) = 12x - 36
  • Step 2: Expand left side. 8x−20=12x−368x - 20 = 12x - 36
  • Step 3: Collect variable terms on right side. −20=12x−8x−36  ⟹  −20=4x−36-20 = 12x - 8x - 36 \implies -20 = 4x - 36
  • Step 4: Add 36 to both sides. 16=4x16 = 4x
  • Step 5: Divide by 4. x=4x = 4

Problem 2: Multi-Step Inequality with Fraction & Negative Coefficient

Problem: Solve the inequality for xx: 7−3x4≤4\frac{7 - 3x}{4} \le 4

Step-by-Step Solution:

  • Step 1: Multiply both sides by +4+4 (positive, so sign does NOT flip). 7−3x≤4⋅47 - 3x \le 4 \cdot 4 7−3x≤167 - 3x \le 16
  • Step 2: Subtract 7 from both sides. −3x≤16−7-3x \le 16 - 7 −3x≤9-3x \le 9
  • Step 3: Divide both sides by −3-3. (DIVIDING BY NEGATIVE: FLIP SIGN from ≤\le to ≥\ge). x≥9−3x \ge \frac{9}{-3} x≥−3x \ge -3
  • Step 4: Conclusion. The solution set is all real numbers greater than or equal to −3-3.
Test Your Knowledge

What is the solution to the linear equation 4(2x - 3) - 2(x + 5) = 3(x - 4)?

A

x = 2

B

x = 4

C

x = 10/3

D

x = -10/3

Test Your Knowledge

Solve for x in the equation containing fractions: (3x - 1)/4 - (x + 3)/2 = 2.

A

x = 7

B

x = 11

C

x = 13

D

x = 15

Test Your Knowledge

Solve the inequality -5x + 14 <= -16 and select the correct solution set.

A

x <= 6

B

x >= 6

C

x <= -6

D

x >= -6

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