Solve for x — Algebra Review

The Golden Rule of Algebra

Whatever you do to one side, do to the other

An equation is a balance scale. As long as you do the same operation to both sides, the equation stays true. Your goal: get x alone on one side.

🎯

The goal: get x by itself

Work backwards through the order of operations. If something was added to x, subtract it. If x was multiplied, divide. Each step undoes one operation until x stands alone.

🔄

Every Operation Has an Opposite

Use the inverse to undo what was done to x

+ addition

↕ undo with

− Subtraction

x + 5 = 12 → x = 12 − 5 = 7

− subtraction

↕ undo with

+ Addition

x − 3 = 8 → x = 8 + 3 = 11

× multiplication

↕ undo with

÷ Division

4x = 20 → x = 20 ÷ 4 = 5

÷ division

↕ undo with

× Multiplication

x6 = 4 → x = 4 × 6 = 24

📋

Order of undoing = reverse order of operations

PEMDAS builds an expression from inside out. To solve, you undo from outside in:
First undo + and − (lowest priority) · Then undo × and ÷ · Then undo exponents/roots

1️⃣

One Operation to Undo

Apply one inverse operation to both sides

x + 9 = 15

1

Undo +9 by subtracting 9 from both sides

2

x + 9 − 9 = 15 − 9

x = 6

3x = 21

1

Undo ×3 by dividing both sides by 3

2

3x/3 = 213

x = 7

x − 4 = 11

1

Undo −4 by adding 4 to both sides

2

x − 4 + 4 = 11 + 4

x = 15

x5 = 8

1

Undo ÷5 by multiplying both sides by 5

2

x5 × 5 = 8 × 5

x = 40

2️⃣

Undo Addition/Subtraction First, Then Multiply/Divide

Always work from the outside in

📌 Example — 3x + 7 = 22

1

Undo +7 first (outermost operation): subtract 7 from both sides → 3x = 15

2

Undo ×3 next: divide both sides by 3 → x = 5

✓

Check: 3(5) + 7 = 15 + 7 = 22 ✅

x = 5

📌 Example — x/4 − 3 = 5

1

Undo −3 first: add 3 to both sides → x/4 = 8

2

Undo ÷4 next: multiply both sides by 4 → x = 32

✓

Check: 324 − 3 = 8 − 3 = 5 ✅

x = 32

📌 Example — −2x + 5 = −9

1

Undo +5: subtract 5 from both sides → −2x = −14

2

Undo ×(−2): divide both sides by −2 → x = 7

✓

Check: −2(7) + 5 = −14 + 5 = −9 ✅

x = 7

📦

Distribute First, Then Solve

Expand parentheses before isolating x

📌 Example — 3(x + 4) = 21

1

Distribute the 3: 3·x + 3·4 = 21 → 3x + 12 = 21

2

Undo +12: subtract 12 → 3x = 9

3

Undo ×3: divide by 3 → x = 3

✓

3(3 + 4) = 3(7) = 21 ✅

x = 3

📌 Example — 2(3x − 5) = 4x + 6

1

Distribute: 6x − 10 = 4x + 6

2

Move x terms to one side: subtract 4x from both sides → 2x − 10 = 6

3

Undo −10: add 10 → 2x = 16

4

Undo ×2: divide by 2 → x = 8

✓

2(3·8−5) = 2(19) = 38 = 4(8)+6 = 32+6 = 38 ✅

x = 8

🍕

Clear the Fractions First — Multiply by the LCD

Eliminate all denominators in one step, then solve normally

💡

Multiply both sides by the LCD (Least Common Denominator)

Find the smallest number that all denominators divide into evenly. Multiply every term on both sides by it. All fractions disappear and you're left with a regular equation.

📌 Example — x3 + x4 = 7

1

LCD of 3 and 4 = 12. Multiply every term by 12.

2

12 · x3 + 12 · x4 = 12 · 7 → 4x + 3x = 84

3

Combine: 7x = 84 → x = 12

✓

123 + 124 = 4 + 3 = 7 ✅

x = 12

📌 Example — 2x − 13 = x + 22

1

LCD of 3 and 2 = 6. Multiply every term by 6.

2

6 · 2x − 13 = 6 · x + 22 → 2(2x−1) = 3(x+2)

3

Distribute: 4x − 2 = 3x + 6

4

Collect x terms: subtract 3x → x − 2 = 6 → x = 8

✓

2·8−13 = 153 = 5 = 8+22 = 102 = 5 ✅

x = 8

🔢

Fraction & Decimal Answers — Totally Normal!

Don't panic when x doesn't come out clean — leave it as a fraction

💡

Fraction answers are exact — decimals can round wrong

When dividing gives a non-integer, leave it as a fraction like 28/3 rather than rounding to a decimal. Fractions are exact. Always simplify by finding the GCF of numerator and denominator.

📌 3x = 28

1

Divide both sides by 3: x = 283

2

28 and 3 share no common factor → already simplified

✓

3 · 283 = 28 ✅

x = 283 (≈ 9.33)

📌 5x + 3 = 14

1

Subtract 3: 5x = 11

2

Divide by 5: x = 115

✓

5 · 115 + 3 = 11 + 3 = 14 ✅

x = 115 (= 2.2)

📌 7x = 42

1

Divide both sides by 7: x = 427 = 6

✓

7(6) = 42 ✅ (This one IS a whole number!)

x = 6

📌 4x − 1 = 6

1

Add 1: 4x = 7

2

Divide by 4: x = 74

✓

4 · 74 − 1 = 7 − 1 = 6 ✅

x = 74 (= 1.75)

↔️

Collect All x Terms on One Side First

Move x terms left, numbers right — then solve as normal

The Strategy

When x appears on both sides, subtract the smaller x term from both sides to gather all x on one side.

1

Move x terms to one side by adding or subtracting the x term from the other side.

2

Move numbers to the other side by adding or subtracting constants.

3

Solve by dividing — you may get a fraction.

📌 Example 1 — 5x + 3 = 2x + 12

1

Subtract 2x from both sides: 5x − 2x + 3 = 12 → 3x + 3 = 12

2

Subtract 3 from both sides: 3x = 9

3

Divide by 3: x = 3

✓

5(3)+3 = 18 = 2(3)+12 = 18 ✅

x = 3

📌 Example 2 — 8x − 4 = 3x + 11

1

Subtract 3x from both sides: 5x − 4 = 11

2

Add 4 to both sides: 5x = 15

3

Divide by 5: x = 3

✓

8(3)−4 = 20 = 3(3)+11 = 20 ✅

x = 3

📌 Example 3 — 7x + 2 = 4x + 9 (fraction answer)

1

Subtract 4x: 3x + 2 = 9

2

Subtract 2: 3x = 7

3

Divide by 3: x = 73

✓

7 · 73+2 = 493 + 63 = 553 = 4 · 73+9 = 283 + 273 = 553 ✅

x = 73

📐

Cross-Multiply to Clear x from the Bottom

When x is in a denominator, cross-multiply first — then solve

✖️

Cross-multiplication: a/b = c/d → ad = bc

When you have a fraction on each side (or can rewrite as one), multiply diagonally: numerator of left × denominator of right = denominator of left × numerator of right. This eliminates all denominators at once.

📌 Example 1 — 5x = 23 (x in denominator on left)

1

Cross-multiply: 5 · 3 = 2 · x → 15 = 2x

2

Divide by 2: x = 152

✓

515/2 = 5 · 215 = 1015 = 23 ✅

x = 152 (= 7.5)

📌 Example 2 — 4x = 6x + 1 (x in denominator on both sides — rewrite first)

1

Rewrite right side with common denominator: 6x + 1 = 6x + xx = 6 + xx

2

Now both sides have denominator x: 4x = 6 + xx

3

Multiply both sides by x: 4 = 6 + x

4

Subtract 6: x = −2

✓

4−2 = −2 = 6−2 + 1 = −3 + 1 = −2 ✅

x = −2

📌 Example 3 — 3x+1 = 2x−1 (x+expression in both denominators)

1

Cross-multiply: 3·(x−1) = 2·(x+1)

2

Distribute: 3x − 3 = 2x + 2

3

Subtract 2x: x − 3 = 2

4

Add 3: x = 5

✓

35+1 = 36 = 12 = 25−1 = 24 = 12 ✅

x = 5

⚠️

Always check x ≠ 0 and denominators ≠ 0

After solving, verify your answer doesn't make any denominator equal zero. If x = 0 makes a denominator zero, it's an extraneous solution — discard it!

√

When the Answer Involves a Square Root

Leave it in exact radical form — don't round

📋

Radical form is exact — decimal is approximate

√3, √2, 2√5 etc. are exact values. A decimal like 1.732… is only an approximation of √3. In algebra and trig, always leave answers in simplified radical form unless told to round.

√4 = 2

Perfect square → whole number

√12 = 2√3

√(4·3) = √4·√3 = 2√3

√18 = 3√2

√(9·2) = √9·√2 = 3√2

📌 Example 1 — x² = 3 (solve for x)

1

Take the square root of both sides: √(x²) = ±√3

2

Remember ± — both positive and negative roots satisfy x² = 3

✓

(√3)² = 3 ✅ (−√3)² = 3 ✅

x = √3 or x = −√3 (i.e. x = ±√3)

📌 Example 2 — 2x² − 6 = 0 (multi-step with radical answer)

1

Add 6 to both sides: 2x² = 6

2

Divide by 2: x² = 3

3

Take square root of both sides: x = ±√3

✓

2(√3)² − 6 = 2(3) − 6 = 0 ✅

x = ±√3 (≈ ±1.732)

📌 Example 3 — 3x² + 1 = 7 (simplify the radical)

1

Subtract 1: 3x² = 6

2

Divide by 3: x² = 2

3

Take square root: x = ±√2

✓

3(√2)² + 1 = 3(2) + 1 = 7 ✅

x = ±√2 (≈ ±1.414)

📌 Example 4 — 5x² = 60 (simplify — pull out the perfect square)

1

Divide by 5: x² = 12

2

Take square root: x = ±√12

3

Simplify √12: √12 = √(4·3) = √4 · √3 = 2√3

✓

5·(2√3)² = 5·(4·3) = 5·12 = 60 ✅

x = ±2√3 (≈ ±3.464)

📌 Example 5 — x√3 = 4 (x involves √3 — rationalize)

1

Multiply both sides by √3: x = 4√3

2

This is already simplified — 4√3 is exact radical form.

✓

4√3√3 = 4 ✅

x = 4√3 (≈ 6.928)

📌 Example 6 — 23 = x · √3 (solve for x — rationalize the answer)

1

Divide both sides by √3: x = 23 ÷ √3 = 23√3

2

Rationalize the denominator — multiply top and bottom by √3 to clear the radical from the bottom:

3

x = 23√3 × √3√3 = 2√33 · (√3 · √3) = 2√33 · 3 = 2√39

4

Check that 2 and 9 share no common factor — GCF(2, 9) = 1, so this is fully simplified.

✓

Verify: x · √3 = 2√39 · √3 = 2 · (√3 · √3)9 = 2 · 39 = 69 = 23 ✅

x = 2√39 (≈ 0.385)

🔑

Rationalizing the denominator

A radical in the denominator (like √3) is not considered fully simplified. To rationalize, multiply top and bottom by that same radical: √3 · √3 = 3 — the radical disappears from the bottom. The value of the fraction doesn't change since you're multiplying by √3√3 = 1.

Equation Type	First Step	Example
x + a = b or x − a = b	Add or subtract a from both sides	x + 5 = 12 → x = 7
ax = b (whole number)	Divide both sides by a	7x = 42 → x = 6
ax = b (fraction answer)	Divide — leave as b/a, simplify if possible	3x = 28 → x = 283
ax + b = c	Subtract b first, then divide by a	2x + 3 = 11 → x = 4
a(x + b) = c	Distribute a, then solve normally	3(x+2) = 15 → x = 3
x terms on both sides	Move all x to one side, numbers to other	5x = 2x + 9 → x = 3
Fractions present	Multiply everything by the LCD first	x2 + x3 = 5 → x = 6
a/x = b/c (x in denominator)	Cross-multiply: a·c = b·x, then solve	5x = 23 → x = 15/2
x² = a (radical answer)	Take √ both sides → ±√a. Simplify radical.	5x² = 60 → x = ±2√3

✅

Always check your answer!

Plug x back into the original equation. If both sides are equal, you're right. If not, you made an arithmetic error somewhere — go back and find it. Checking takes 10 seconds and catches everything.