Linear Equations & Inequalities

A1.B.1Multi-step equations01/05

Solve multi-step linear equations.

Solve for x

3 (2 x - 1) + 4 = 5 x + 7

Step01of 05

Step 1 — distribute first. Clear any parentheses before combining anything.

3 (2 x - 1) + 4 = 5 x + 7

6 x - 3 + 4 = 5 x + 7

A1.B.1 — Solve multi-step linear equations.

Solve for x

3 (2 x - 1) + 4 = 5 x + 7

Step 01 of 05

Step 1 — distribute first. Clear any parentheses before combining anything.

3 (2 x - 1) + 4 = 5 x + 7

6 x - 3 + 4 = 5 x + 7

Step 02 of 05

Step 2 — combine like terms on each side. Tidy up before moving anything across the equals sign.

6 x + 1 = 5 x + 7

Step 03 of 05

Step 3 — get the variable on one side. Subtract $5 x$ from BOTH sides — whatever you do to one side, do to the other.

6 x - 5 x + 1 = 5 x - 5 x + 7

x + 1 = 7

Step 04 of 05

Step 4 — isolate the variable. Subtract 1 from both sides.

x = 6

Step 05 of 05

Step 5 — always check. Plug $x = 6$ back into the ORIGINAL.

LHS: RHS: 3 (2 \cdot 6 - 1) + 4 = 3 (11) + 4 = 37 5 (6) + 7 = 37 ✓

Key insight

The order is fixed: distribute → combine → isolate the variable → divide. Skipping the distribution step is the #1 source of wrong answers in multi-step equations.

A1.B.2 — Equations with variables on both sides.

Solve for x

5 x - 3 = 2 x + 12

Step 01 of 05

When the variable appears on BOTH sides, the goal stays the same: get all variable terms on one side, all constants on the other.

Pick the side with the SMALLER coefficient to subtract from. Here that's $2 x$ on the right. Subtracting it keeps the remaining variable coefficient positive (less sign trouble).

Step 02 of 05

Subtract $2 x$ from both sides.

5 x - 2 x - 3 = 2 x - 2 x + 12

3 x - 3 = 12

Step 03 of 05

Now it's a single-variable equation. Move the constant: add 3 to both sides.

3 x = 15

Step 04 of 05

Divide by the coefficient.

x = 5

Verify: $5 (5) - 3 = 22$ and $2 (5) + 12 = 22$ $✓$ .

Step 05 of 05

Special outcomes. Sometimes the variable cancels entirely.

$5 = 5$ (true)	→infinitely many solutions
$5 = 7$ (false)	→no solution
$x = c$	→one solution

Key insight

Subtract the smaller-coefficient variable term first to keep things positive. If the variable disappears, read the resulting numeric statement — true means all reals work, false means nothing works.

A1.B.3 — Literal equations — solve for one variable in terms of others.

Solve for h

A = \frac{1}{2} bh

Step 01 of 05

A literal equation uses letters instead of numbers. The mechanics are identical to a numeric equation — undo each operation around the target variable until it stands alone.

The triangle area formula $A = \frac{1}{2} bh$ is normally used to find $A$ . To use it to find the HEIGHT given area and base, we rearrange.

Step 02 of 05

Step 1 — clear the fraction. Multiply both sides by 2.

2 A = bh

Equivalent move: multiply both sides by 2. The fraction was the outermost wrapper around $bh$ , so it comes off first.

Step 03 of 05

Step 2 — divide by everything else attached to the target. $h$ is multiplied by $b$ , so divide both sides by $b$ .

\frac{2 A}{b} = h

h = \frac{2 A}{b}

Step 04 of 05

Verify with numbers. If $A = 30$ and $b = 10$ :

h = \frac{2 ( 30 )}{10} = 6

Check: $A = \frac{1}{2} (10) (6) = 30$ $✓$ .

Step 05 of 05

One more example — solve $P = 2 L + 2 W$ for $W$ .

P - 2 L W = 2 W = \frac{P - 2 L}{2}

Key insight

Think of literal equations as solving with letters in place of numbers. Same moves, same order: clear fractions, undo addition/subtraction, undo multiplication/division. Target variable ends alone on one side.

A1.B.4 — Compound inequalities — AND vs. OR.

Step 01 of 05

A compound inequality joins two inequalities with AND or OR.

AND	→both must hold — solution is the overlap
OR	→either may hold — solution is the union
"between"	→almost always AND
"less than … or greater than …"	→almost always OR

Step 02 of 05

AND example. Solve $- 3 < 2 x + 1 \leq 7$ — the compact "between" form. Apply every operation to ALL THREE parts.

- 3 - 1 - 4 - 2 < 2 x + 1 - 1 \leq 7 - 1 < 2 x \leq 6 < x \leq 3

Step 03 of 05

The solution is the segment between $- 2$ and $3$ , with an open dot at $- 2$ (strict $<$ ) and a closed dot at $3$ ( $\leq$ ).

- 2 < x \leq 3

Step 04 of 05

OR example. Solve $x - 1 < - 3 or x - 1 \geq 4$ . Solve each piece independently.

x < - 2 or x \geq 5

The solution is two rays pointing outward.

Two rays — open at

- 2

, closed at

5

Step 05 of 05

The trap with negatives. Multiplying or dividing by a negative number FLIPS each inequality sign.

- 2 x < 6 \Rightarrow x > - 3 (not x < - 3)

Key insight

AND = overlap (one segment between two values). OR = union (two rays going outward). When you flip signs by dividing by a negative, flip BOTH inequalities in a compound statement, not just one.

A1.B.5 — Absolute value equations.

Solve for x

∣ x - 3∣ = 7

Step 01 of 05

$∣ x ∣$ is the distance from x to $0$ on the number line. So $∣ x - 3∣ = 7$ says: "the distance from x to $3$ is exactly $7$ ." Two values do that — one to the left, one to the right.

Step 02 of 05

Split into two equations. The expression inside the bars equals either $7$ OR $- 7$ .

x - 3 = 7 or x - 3 = - 7

Step 03 of 05

Solve each.

x = 10 or x = - 4

Both check (distances 7 from 3) — solution set ${- 4, 10}$ .

Both

- 4

and

10

sit 7 units from

3

Step 04 of 05

Always isolate the absolute value first. If the equation reads $2∣ x - 3∣ + 1 = 15$ , peel off the +1 and the $2$ before splitting:

2∣ x - 3∣ + 1 2∣ x - 3∣ ∣ x - 3∣ = 15 = 14 = 7

NOW split into two cases. Skipping this step is the most common error.

Step 05 of 05

When there's NO solution. Absolute value is never negative, so:

∣ x - 3∣ = - 5 \Rightarrow no solution

If you isolate the absolute-value expression and find it equals a negative number, stop. There's nothing to solve.

Key insight

Two-step recipe: isolate

∣...∣

→ split into the two cases (= positive and = negative) → solve each. Negative = no solution.

A1.B.6 — Absolute value inequalities — "less than" vs. "greater than".

Step 01 of 05

Absolute value inequalities split into two cases just like equations, but the SHAPE of the solution depends on the inequality type.

$∣ x ∣ < c$ ("less thAND")	→AND — single segment between $- c$ and $c$
$∣ x ∣ > c$ ("greatOR")	→OR — two rays outside $- c$ and $c$

The mnemonic: "less thAND, greatOR." Memorize once; the mistake disappears forever.

Step 02 of 05

"Less than" example. Solve $∣ x - 3∣ < 7$ . Rewrite as the AND form (between):

- 7 < x - 3 < 7

Add 3 to all three parts:

- 4 < x < 10

One segment, both endpoints open

Step 03 of 05

"Greater than" example. Solve $∣ x - 3∣ \geq 7$ . Rewrite as the OR form:

x - 3 \leq - 7 or x - 3 \geq 7

Add 3 to each piece:

x \leq - 4 or x \geq 10

Two rays, both endpoints closed

Step 04 of 05

The same trap as equations: isolate the absolute value first, and flag negatives.

∣ x ∣ < - 5 \Rightarrow no solution

∣ x ∣ > - 5 \Rightarrow all reals

Absolute value is never negative, so the first is impossible (nothing works) and the second is automatic (everything works).

Step 05 of 05

Why "less thAND, greatOR" is the only thing to memorize. Absolute value measures distance. "Distance less than 7" puts you in a band around the center — one segment. "Distance more than 7" puts you outside the band — two rays. The shape comes from the geometry, not the algebra.

Key insight

Less than → AND → one segment. Greater than → OR → two rays. Always isolate the absolute value before splitting; if it ends up less than a negative, no solution.

Free diagnostic

Test your understanding: 10 questions, ~10 min.

Pulled live from our bank for A1.B · scored instantly with worked solutions · free, no sign-up.

Start diagnostic →Print worksheets

A1.B.1 – A1.B.6 · CCSS A.REI.B.3 / A.CED.A.4 / A.REI.D.11·Full curriculum →