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The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 2 (Intermediate Algebra) Solving Linear Equations

Overview

This lecture is a thorough tour of solving linear equations — equations in which the variable appears only to the first power. We start from the simplest one-step cases and build up systematically: subtracting a constant, dividing by a coefficient, distributing, combining like terms, collecting variables onto one side, and clearing fractions. The unifying idea is isolating the variable: whatever you do to one side of an equation, you must do to the other, until the variable stands alone. The final segment generalizes the same moves to literal equations, where you solve for one letter in terms of others (e.g. rearranging a formula). Mastering this routine is the foundation for nearly everything that follows in algebra.

Key Concepts

What makes an equation linear

The two inverse operations

General form and recipe

Sign rule for products

Order of operations when solving

  1. Distribute to clear parentheses: a(x−y)=ax−aya(x - y) = ax - ay.
  2. Combine like terms on each side.
  3. Collect all variable terms on one side and all constants on the other (usually move the variable to the side that already has more of it, to avoid a negative coefficient).
  4. Isolate the variable by dividing by its coefficient.

Handling a negative variable term

Fractions don't change the method

Literal equations (solving for a variable)

Worked Example

Solve a multi-step linear equation with variables on both sides:

4(2x−6)+9=3(x−7)+8x8x−24+9=3x−21+8x8x−15=11x−218x−15−8x=11x−21−8x−15=3x−21−15+21=3x6=3xx=2\begin{aligned} 4(2x - 6) + 9 &= 3(x - 7) + 8x \\ 8x - 24 + 9 &= 3x - 21 + 8x \\ 8x - 15 &= 11x - 21 \\ 8x - 15 - 8x &= 11x - 21 - 8x \\ -15 &= 3x - 21 \\ -15 + 21 &= 3x \\ 6 &= 3x \\ x &= 2 \end{aligned}

Check by substituting x=2x = 2: the left side gives 4(4−6)+9=4(−2)+9=14(4-6)+9 = 4(-2)+9 = 1, and the right side gives 3(2−7)+16=3(−5)+16=13(2-7)+16 = 3(-5)+16 = 1. Both sides agree, so x=2x = 2 is correct.

For a literal-equation example, solving A=πr2+πrsA = \pi r^2 + \pi r s for ss:

A−πr2=πrss=A−πr2πr\begin{aligned} A - \pi r^2 &= \pi r s \\ s &= \frac{A - \pi r^2}{\pi r} \end{aligned}

Summary