Linear Equations

The balance metaphor is literally true

When Bedrock built the real numbers, it established two critical facts about equations: (1) if $a = b$ then $a + c = b + c$ for any $c$ (add the same thing to both sides); (2) if $a = b$ and $c \ne 0$ then $ac = bc$ (multiply both sides by the same nonzero thing). These are not magic rules for algebra class — they are theorems about equality and the real-number operations. Solving an equation is applying these two facts strategically to peel away everything around the variable until it stands alone.

The variable is the masked number from the Variables and Expressions lesson. Your job is to unmask it.

Solving $ax + b = c$: the canonical recipe

Consider $3x + 7 = 19$.

Goal: get $x$ alone on the left. It is surrounded by "multiply by 3" and "add 7". Undo them in reverse order — last operation on, first operation off.

Step 1. Undo "add 7" by adding $-7$ to both sides: $3x + 7 + (-7) = 19 + (-7) \implies 3x = 12.$

Step 2. Undo "multiply by 3" by multiplying both sides by $\frac{1}{3}$: $\frac{1}{3} \cdot 3x = \frac{1}{3} \cdot 12 \implies x = 4.$

Step 3 (mandatory). CHECK by substituting back: $3(4) + 7 = 12 + 7 = 19. \checkmark$

The check is a free correctness proof. It takes ten seconds and catches every arithmetic slip. Use it every single time.

Variables on both sides

When variables appear on both sides, collect them first. Suppose $5x - 3 = 2x + 9$.

Subtract $2x$ from both sides (legal — same thing to both sides): $5x - 2x - 3 = 9 \implies 3x - 3 = 9.$

Now it looks like the previous pattern: $3x = 12 \implies x = 4.$

Check: $5(4) - 3 = 17$ and $2(4) + 9 = 17$. Both sides equal 17. Done.

Clearing fractions and parentheses

Fractions inside equations are annoying. The cure: multiply both sides by the LCD before doing anything else. The Fractions and Rationals lesson established that multiplying by a nonzero number preserves equality — use it.

$\frac{x}{3} + \frac{x}{4} = 7.$

LCD of 3 and 4 is 12. Multiply every term by 12: $12 \cdot \frac{x}{3} + 12 \cdot \frac{x}{4} = 12 \cdot 7 \implies 4x + 3x = 84 \implies 7x = 84 \implies x = 12.$

Check: $\frac{12}{3} + \frac{12}{4} = 4 + 3 = 7$. Confirmed.

Parentheses get distributed first — that's distributivity from Bedrock: $2(3x - 4) = 5x + 1 \implies 6x - 8 = 5x + 1 \implies x = 9.$

Check: $2(3(9)-4) = 2(23) = 46$ and $5(9)+1 = 46$. Confirmed.

Special cases: two flavors of drama

Case 1: Identity. Solve $3(x + 2) = 3x + 6$.

Expand: $3x + 6 = 3x + 6$. Subtract $3x$ from both sides: $6 = 6$.

No $x$ in sight — and $6 = 6$ is TRUE for every value of $x$. Every real number is a solution. The equation is an identity. Algebraic symptom: a true numeric statement with no variables remaining.

Case 2: Contradiction. Solve $3(x + 2) = 3x + 11$.

Expand: $3x + 6 = 3x + 11$. Subtract $3x$: $6 = 11$.

FALSE for every value of $x$. No solution exists. The equation is a contradiction. Algebraic symptom: a false numeric statement with no variables remaining.

These are not errors — they are the equation's structure announcing itself.

Literal equations

Solve a formula for one of its variables by treating every other letter as a constant. The legal moves are identical.

Solve $A = \frac{1}{2}bh$ for $h$: $A = \frac{1}{2}bh \implies 2A = bh \implies h = \frac{2A}{b}, \quad b \ne 0.$

Same two moves. Different labels on the constants.