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Mathematics 16 Online
OpenStudy (anonymous):

Step by step to answering |8*x−1| ≤2*x+11.

OpenStudy (anonymous):

\[|x| < a \rightarrow -a < x < a\] Use it and solve

OpenStudy (kc_kennylau):

Entonces, tenemos: \[-(2x+11)\le8x-1\le2x+11\]

OpenStudy (kc_kennylau):

Porque el "x" en su ecuacion es "8x-1", y el "a" en su ecuacion es "2x+11"

OpenStudy (anonymous):

????

OpenStudy (kc_kennylau):

@sourwing ?

OpenStudy (anonymous):

creo que si entiendo un poco

OpenStudy (anonymous):

is this spanish?

OpenStudy (anonymous):

yes

OpenStudy (kc_kennylau):

@AntonCosme si aun no entiendes, puede lo pensar como asi: \[|y|<a\rightarrow -a<y<a\]

OpenStudy (kc_kennylau):

Si no usas "x", lo puedes entender

OpenStudy (kc_kennylau):

entonces, "y" significa 8x-1, y "a" significa 2x+11

OpenStudy (anonymous):

Si lo entiendo mas mejor.

OpenStudy (kc_kennylau):

no lo entiendes?

OpenStudy (anonymous):

Si entindi como ponerlo en la formula que usted escribio pero no se como completar el problema

OpenStudy (kc_kennylau):

tenemos \(−(2x+11)≤8x−1≤2x+11\) Los puedes separar a: \(-2x-11\le8x-1\) y \(8x-1\le2x+11\)

OpenStudy (anonymous):

Ok

OpenStudy (kc_kennylau):

Solve las dos ecuaciones separadamente Disculpe, tengo que comer ahorita

OpenStudy (kc_kennylau):

Cual huso horario eres en?

OpenStudy (kc_kennylau):

GMT+8 para mi

OpenStudy (anonymous):

esta bien tengo que hirme a dormir es tarme 12:44AM Tampa Fl

OpenStudy (kc_kennylau):

Entonces es GMT-5?

OpenStudy (kc_kennylau):

Hasta manana :D

OpenStudy (anonymous):

si gracias por la ayuda.

OpenStudy (kc_kennylau):

(u hoy)

OpenStudy (kc_kennylau):

adios

OpenStudy (anonymous):

bye

OpenStudy (kc_kennylau):

de nada :)

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