Linear Equations in One Variable

Run the machine backwards

In F8 you built a phone-bill machine: at $$20$ a month plus $$3$ per gigabyte, the bill is $20 + 3g$. Feed it a usage, out comes a bill. Now flip the problem: the bill arrives at $$35$ — how many gigabytes did you use? This time you know the output and want the input, and writing that down gives you your first equation:

$20 + 3g = 35$

Solving it means finding the value of $g$ that makes both sides genuinely equal. All of algebra’s most famous machinery — this module — exists to answer questions shaped like that.

The one rule: keep the scale balanced

An equation is a balance scale that hangs perfectly level: $20 + 3g$ sits in one pan, $35$ in the other. You’re allowed to do anything to the scale that keeps it level, which means: whatever you do to one side, you must do to the other. Add $3$ to both pans — still level. Add $3$ to one pan only, and the scale tips; $5 + 3 \ne 5$. Every legal solving move is just this one rule applied with a purpose.

So let’s find the gigabytes, one balanced move at a time:

Notice the order of the undoing. Building the bill, the machine multiplied first and added second (F1’s ladder). Freeing $g$, you undid the addition first and the multiplication second — inverse operations, in reverse order, like taking off shoes before socks:

“Move it to the other side” — what’s actually happening

You’ll hear solving described as moving terms: “the $+4$ moves over and becomes $-4$.” That shorthand is fine once you know what it hides, and dangerous before. Nothing moves. In $3x + 4 = 19$, you subtract $4$ from both sides; on the left it cancels to nothing, on the right it surfaces as $19 - 4$. The “sign flip” isn’t a rule about crossing the equals sign — it’s the visible leftover of a both-sides move. People who memorize it as teleportation eventually flip a sign that shouldn’t flip (or drag a coefficient across as if it worked the same way). When in doubt, fall back to the scale: name the operation, do it to both pans.

Variables on both sides

$5x + 4 = 2x + 19$ looks new, but $2x$ is just a quantity like any other — so subtract $2x$ from both pans and it’s gone from the right: $3x + 4 = 19$. From there you know the drill. (Collecting the smaller $x$-term keeps the coefficient positive, which is kinder to your signs.)

The solver opens on that exact equation. Before reading each step, predict the move: which $x$-term gets collected, what gets undone first, and what’s the final $x$? Then type an equation with parentheses — say $4 - 2(x - 3) = 8$ — and watch the F8 skills (distribute, watch the negative!) become the opening moves of solving.

When x vanishes: no solution or infinitely many

Try $3(x + 2) = 3x + 6$ in the solver, then $2x + 5 = 2x + 1$. In both, the $x$-terms cancel completely — and what’s left tells you which strange case you’re in. $6 = 6$ is true no matter what $x$ was: the two sides were the same machine in different clothes, so every number solves it — infinitely many solutions. $5 = 1$ is false no matter what: the two sides always differ by exactly $4$, so no number can ever reconcile them. Neither is an error — both are answers, and the SAT loves asking you to recognize them.

Always check

Plug your answer back into the original equation — not a later line, which may already carry your mistake — and confirm both sides land on the same number. It’s the same five-second lie detector from F8, and it catches nearly every slip. Build the habit in the Check a Solution tab.

The one thing to remember

An equation is a level scale, and solving is undoing: peel everything off $x$ with inverse operations, in reverse order of operations, doing each move to both sides. “Moving terms” is only ever a both-sides move wearing a nickname — and a plugged-back-in answer never lies.

What a linear equation is

An equation says two things are equal: a left side and a right side joined by $=$. A linear equation has the variable to the first power only — no $x^2$, no $x$ in a denominator. Solving means finding the value of $x$ that makes both sides truly equal.

The strategy (same every time)

Work backwards through the order of operations to peel $x$ free:

1. Clear parentheses — distribute.
2. Clear fractions — multiply every term by the common denominator (optional but tidy).
3. Collect like terms — all $x$-terms on one side, all constants on the other.
4. Undo what’s done to $x$ — subtract/add the constant, then divide by the coefficient.

Each move uses an inverse operation: addition undoes subtraction, multiplication undoes division.

Worked example — two-step

$3x - 7 = 11$

Worked example — variables on both sides

Get the $x$‘s together first. Move the smaller $x$-term to avoid a negative if you can.

$5x + 4 = 2x + 19$

Worked example — fractions

Multiply every term by the common denominator to clear them.

$\frac{x}{2} + \frac{1}{3} = 2$

Two special answers

Sometimes the $x$‘s cancel out completely. Look at what’s left:

The SAT loves to ask “for what value does this have no / infinitely many solutions” — recognizing these is worth points.

Always check

Plug your answer back into the original equation. If both sides match, you’re right. This single habit catches almost every arithmetic slip.