Solving Linear Equations

What a Linear Equation Is

A linear equation in one variable is any equation you can rewrite in the form

$ax + b = c$

where $a$, $b$, and $c$ are constants and the variable $x$ appears only to the first power — no $x^2$, no $\sqrt{x}$, no $x$ in a denominator. Because the graph of such a relationship is a straight line, "linear" equations always have exactly one solution unless the variable terms cancel completely, a special case the SAT loves to test.

Your job is almost always the same: isolate the variable. Every legal move you make is aimed at getting $x$ alone on one side. This is the most frequently tested skill on SAT Math because it is the engine underneath systems, inequalities, and word problems — get fluent here and the harder topics get much easier.

The Balance Principle

Think of an equation as a balanced scale: the left side weighs exactly as much as the right side. You may do anything to the equation as long as you do the same thing to both sides. Add 5 to the left, add 5 to the right. Divide the left by 3, divide the right by 3. The scale stays balanced, so the solution never changes.

That single rule justifies the whole procedure:

1. Simplify each side separately — distribute across parentheses and combine like terms.
2. Collect the variable on one side — add or subtract a variable term from both sides.
3. Collect the constants on the other side — add or subtract a number from both sides.
4. Divide by the coefficient — turn $ax = d$ into $x = \tfrac{d}{a}$.

A quick model solve

Solve $3x + 5 = 14$:

$\begin{aligned} 3x + 5 &= 14 \\ 3x &= 9 \\ x &= 3 \end{aligned}$

Subtracting 5 undoes the $+5$; dividing by 3 undoes the "times 3." Notice the operations are applied in reverse order of operations — you peel the constant off before the coefficient.

Variables on Both Sides

When $x$ appears on the left and the right, first move all the variable terms to one side. Solve $5y - 3 = 2y + 9$ by subtracting $2y$ from both sides so the variable lives on the left only:

$\begin{aligned} 5y - 3 &= 2y + 9 \\ 3y - 3 &= 9 \\ 3y &= 12 \\ y &= 4 \end{aligned}$

It does not matter which side you collect the variable on — moving $5y$ to the right gives $-3 = -3y + 9$, then $y = 4$ again. Pick the side that keeps the coefficient positive; it prevents sign mistakes.

Distributing and Clearing Fractions

Two setups slow students down: parentheses and fractions.

• Parentheses: distribute first. In $4(x - 3) + 2 = 3x - 1$, expand to $4x - 12 + 2 = 3x - 1$, simplify the left to $4x - 10$, then solve. The most common error here is forgetting to multiply the second term inside the parentheses.
• Fractions: multiply every term on both sides by the least common denominator to clear them in one stroke. For $\tfrac{x}{3} + \tfrac{x}{4} = 7$, multiply through by 12: $4x + 3x = 84$, so $7x = 84$ and $x = 12$. Clearing denominators up front is far safer than juggling fractions step by step.

The Special Cases the SAT Tests

Because a linear equation is really a comparison of two lines, three outcomes are possible after you simplify:

The SAT dresses this up with a parameter. "For what value of $b$ does $2(3x - 4) = 6x + b$ have infinitely many solutions?" Expand the left to $6x - 8$. The equation becomes $6x - 8 = 6x + b$. The $6x$ terms cancel, leaving $-8 = b$. So $b = -8$ makes the two sides identical (infinitely many solutions); any other value of $b$ leaves a contradiction and gives no solution. Recognizing that the $x$-terms are about to cancel — because the coefficients already match — is the whole insight.

Always Check

Substituting your answer back into the original equation catches almost every arithmetic slip. It costs ten seconds and turns a guess into a certainty, which is exactly the trade-off you want on a timed test.