If x-terms show up on both sides, treat one x-term like a plain number and subtract it from both sides first.
Si aparecen términos con x en ambos lados, trata uno de ellos como un número común y réstalo de ambos lados primero.
| 6(x + 1) − 2x − 1 = (x + 15) + (x + 16) | Starting equation |
| 6x + 6 − 2x − 1 = x + 15 + x + 16 | Distribute the 6Distribuye el 6 |
| 4x + 5 = 2x + 31 | Combine like terms on each sideCombina términos semejantes en cada lado |
| 2x + 5 = 31 | Subtract 2x from both sidesResta 2x de ambos lados |
| x = 13 | Subtract 5, then divide by 2Resta 5, luego divide entre 2 |
More than one letter — solve for one, treating the others as numbers.
Más de una letra — resuelve para una, tratando las demás como números.
| A = ½bh → h = 2A/b | Multiply both sides by 2, then divide by bMultiplica ambos lados por 2, luego divide entre b |
Multi-step equations are the same 'unbuild the house' idea, just with a cleanup step first: distribute anything stuck outside parentheses, combine like terms on each side, and get all x-terms on one side. Then it's a two-step equation you already know how to solve.
| Solve: 5x − 2(x + 4) = 10 | Distribute the −2: 5x − 2x − 8 = 10 → 3x − 8 = 10 → 3x = 18 → x = 6 |
| Solve: 4x + 9 = 7x − 6 | x on both sides: subtract 4x → 9 = 3x − 6 → 15 = 3x → x = 5. Moving the smaller x-term keeps the coefficient positive. |
"And" needs both parts true. "Or" needs only one.
"Y" necesita que ambas partes sean verdaderas. "O" necesita solo una.
Bracket [ ] = endpoint included. Parenthesis ( ) = endpoint excluded.
Corchete [ ] = el extremo está incluido. Paréntesis ( ) = el extremo está excluido.
| −8 < x ≤ 10 → (−8, 10] | Parenthesis on −8 (excluded), bracket on 10 (included)Paréntesis en −8 (excluido), corchete en 10 (incluido) |
| −2x + 8 < 12 | Starting | |
| −2x < 4 | Subtract 8 — no flipResta 8 — no hay cambio | |
| x > −2 | Divide by −2 — FLIPDivide entre −2 — SE INVIERTE |
An AND compound inequality is a value trapped between two fences — whatever you do to the middle, do to all three parts. An OR inequality is two separate problems whose answers you simply keep together. The one trap: multiplying or dividing by a negative flips every inequality symbol.
| Solve: −5 ≤ 3x + 1 < 10 | Subtract 1 from all three parts: −6 ≤ 3x < 9. Divide all by 3: −2 ≤ x < 3. |
| Solve: −2x + 3 > 11 | Subtract 3: −2x > 8. Divide by −2 and FLIP: x < −4. |
Let n = smallest. The next two are n+1 and n+2.
Sea n = el menor. Los siguientes dos son n+1 y n+2.
| n + (n+1) + (n+2) = 2(n+2) + 12 | "Sum of three" = twice the largest, plus 12"Suma de los tres" = el doble del mayor, más 12 |
| 3(n + 5) ≥ 17 | "Three times the sum of n and 5 is at least 17" — parentheses matter"Tres veces la suma de n y 5 es por lo menos 17" — los paréntesis importan |
| 8.50n + 25 ≥ 122 → n ≥ 11.4 | Round UP to 12 weeks — 11 alone isn't enoughRedondea HACIA ARRIBA a 12 semanas — 11 solas no alcanzan |
Modeling is translation, not new math. Circle the quantity that changes (that's your variable), find the flat/starting amount, and find the per-unit rate. 'Is' means equals; 'at least' means ≥; 'at most' means ≤. Write the sentence in symbols, then solve like any equation.
| A phone costs $50 plus $0.10 per text. Total bill: $62. How many texts? | 50 + 0.10t = 62 → 0.10t = 12 → t = 120 texts. The $50 is the flat part; $0.10 rides with the variable. |
| Three consecutive integers sum to 48. Find them. | n + (n+1) + (n+2) = 48 → 3n + 3 = 48 → n = 15, so 15, 16, 17. |
These problems pull from earlier units on purpose. Switching between skills feels harder in the moment, but it helps you remember longer and matches how a real exam mixes topics. Try each one on paper first, then reveal.Estos problemas provienen de unidades anteriores a propósito. Cambiar de una destreza a otra se siente más difícil en el momento, pero te ayuda a recordar por más tiempo y se parece a cómo un examen real mezcla los temas. Intenta cada uno en papel primero, luego revela la respuesta.