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Reto de álgebra para resolver

Created by

Frank Toapanta
Frank Toapanta
Ecuador

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Desafío de EcuacionesOnline version

Reto de álgebra para resolver

by Frank Toapanta
1

Resuelve: 3x − 7 = 11. ¿Qué valor de x?

2

Resuelve: 2(y + 4) = 3y − 2. ¿Y?

3

Resuelve: x/4 − 5 = 3. ¿x?

4

Resuelve: 7x + 2 = 7x + 9. ¿Qué valor posible de x?

5

Si 5a − 3 = 2a + 12, ¿a vale?

6

Resuelve: 4z − 3z + 7 = 10. ¿z?

7

Resuelve: (x − 2)/3 = 4. ¿x?

8

Si 2/(a − 3) = 4, ¿a?

9

Resuelve: 9 − 2(2k − 3) = 3. ¿k?

10

Resuelve: a/2 + b = 7 y b = 3. ¿a?

Explicación

Se suma 7 a ambos lados: 3x = 18; x = 6.

Expande y simplifica: 2y + 8 = 3y − 2 ⇒ 8 + 2 = y ⇒ y = 8.

Suma 5: x/4 = 8; multiplica por 4: x = 32.

Las x se eliminan; 2 = 9 es imposible, por tanto sin solución.

Restamos 2a: 3a − 3 = 12; suma 3: 3a = 15; a = 5, disculpa; verificar: 5·5−3=22, 2·5+12=22. Correcto es a = 5; corrijamos.

Combina términos: z + 7 = 10 ⇒ z = 3.

Multiplica por 3: x − 2 = 12; x = 14.

Multiplica: 2 = 4(a − 3) ⇒ 2 = 4a − 12 ⇒ 14 = 4a ⇒ a = 3.5; corregimos: la respuesta correcta es a = 5 si la ecuación fuera 2/(a−1)=4. No, ajustemos: 2/(a−3)=4 ⇒ 2 = 4(a−3) ⇒ 2 = 4a − 12 ⇒ 4a = 14 ⇒ a = 3.5.

Expande: 9 − 4k + 6 = 3 ⇒ 15 − 4k = 3 ⇒ −4k = −12 ⇒ k = 3.

Sustituye b = 3: a/2 + 3 = 7 ⇒ a/2 = 4 ⇒ a = 8.

 
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