6.3 Solving Linear Equations

Solving equations is the payoff for everything in sections 6.1 and 6.2. Paper D (Year 6) uses number sentences with all four operations; Papers E-F (Years 7-8) list linear equations explicitly; Papers G-H (Years 9-10) add solving 'graphically and algebraically'. Since personal calculators are not allowed, every step is mental or written arithmetic, so the balance method - small, reversible moves - is your best friend.

The Balance Principle

An equation states that two expressions are EQUAL. Picture a set of balance scales: the left pan and right pan hold the same weight. You may add, subtract, multiply or divide - but whatever you do to one side, you must do to the other, or the scales tip.

One-step equations

Undo the single operation with its inverse:

• x + 7 = 15: subtract 7 from both sides, so x = 8.
• 4x = 36: divide both sides by 4, so x = 9.
• x/5 = 6: multiply both sides by 5, so x = 30.

Two-step equations

Undo operations in REVERSE order: deal with the addition or subtraction first, then the multiplication or division.

Worked example: 3x - 5 = 16.

• Add 5 to both sides: 3x = 21.
• Divide both sides by 3: x = 7.
• Check: 3(7) - 5 = 21 - 5 = 16. Correct.

The order matters. Dividing first (3x - 5)/3 = 16/3 makes the arithmetic far messier - undo the '- 5' before the 'x 3'.

Equations with Brackets

Expand the brackets first, then solve as before.

Worked example: 4(x + 3) = 30.

• Option A (divide first): x + 3 = 7.5, then x = 4.5 - valid when the right side divides evenly by 4, which 30 does not. So prefer Option B.
• Option B (expand first): 4x + 12 = 30, then 4x = 18, so x = 4.5.

Worked example: 5(2x - 1) = 3x + 16.

• Expand: 10x - 5 = 3x + 16.
• Subtract 3x from both sides: 7x - 5 = 16.
• Add 5: 7x = 21, so x = 3. Check: left side 5(6 - 1) = 25; right side 9 + 16 = 25. Correct.

Unknowns on Both Sides

The goal is to collect all variable terms on one side and all constants on the other. Move the SMALLER variable term first to keep coefficients positive and avoid sign errors.

Worked example: 7x - 4 = 3x + 20.

• Subtract 3x: 4x - 4 = 20.
• Add 4: 4x = 24, so x = 6.

Worked example with negatives: 5x + 9 = 2x - 3.

• Subtract 2x: 3x + 9 = -3.
• Subtract 9: 3x = -12, so x = -4. Negative answers are legitimate - check: 5(-4) + 9 = -11 and 2(-4) - 3 = -11. Correct.

Equations with Fractions

Multiply EVERY term on both sides by the denominator to clear the fraction. For x/3 + 2 = 7, multiply everything by 3: x + 6 = 21, so x = 15. With two denominators, use the lowest common denominator: for x/2 + x/3 = 10, multiply by 6 to get 3x + 2x = 60, so 5x = 60 and x = 12. A classic trap is multiplying only the fraction terms and forgetting a standalone constant - every term must be multiplied.

Checking by Substitution

Substitute your answer back into the ORIGINAL equation (not a line from your working, which could contain an error). Both sides must produce the same number. This takes under a minute and catches nearly every slip; on a multiple-choice paper you can also work backwards from the options by substituting each candidate value until one balances the equation - a completely legitimate ICAS strategy.

Forming Equations from Word Problems

ICAS loves 'think of a number' and perimeter/money problems:

• 'A number is multiplied by 4 and then 7 is added; the result is 39.' Form: 4n + 7 = 39, so 4n = 32 and n = 8.
• 'A rectangle's length is 3 cm more than its width; its perimeter is 26 cm.' Let width = w, length = w + 3. Perimeter: 2w + 2(w + 3) = 26, so 4w + 6 = 26, w = 5 cm (and length 8 cm).

The routine: define the variable in words, translate each fact into an expression, build the equation, solve, then answer the question ACTUALLY asked (often the length, not w).

Solving Graphically vs Algebraically (Papers G-H)

Algebraically, 2x + 1 = 7 gives x = 3 in two moves. Graphically, you plot y = 2x + 1 and y = 7 on the cartesian plane; the x-coordinate of the intersection point is the solution. Both lines cross at (3, 7), confirming x = 3. ICAS may show a graph of two lines and ask which equation's solution the intersection represents, or ask you to read an approximate solution from a graph when the intersection is not on a grid line. Key idea: the graphical method solves ANY equation of the form f(x) = g(x) by finding where the two graphs meet, while the algebraic method is exact and usually faster for linear equations.

Quick-reference: inverse operations