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Please help I only have until 8:15 AZ time ;-;
Twice a​ number, increased by one​, is between negative 3 and 9. Find all the numbers.


Sagot :

Step-by-step explanation:

-3 < 2x+1 <9

-3-1 < 2x < 9-1

-4 < 2x < 8

-2 < x < 4

so, the numbers are between -2 and 4

for Integers : -1, 0, 1, 2, 3

[tex]\large \mathfrak{Solution : }[/tex]

let the number be " x "

The Inequality can be represented as :

  • [tex] - 3 < 2x + 1 < 9[/tex]

  • [tex] - 3 - 1 < 2x + 1 - 1 < 9 - 1[/tex]

  • [tex] - 4 < 2x < 8[/tex]

  • [tex] \dfrac{ - 4}{2} < \dfrac{2x}{2} < \dfrac{8}{2} [/tex]

  • [tex] - 2 < x < 4[/tex]

therefore the number actually lies between - 2 and 4, so the possible values for the number (if it's an integer) is -1 , 0 , 1 , 2 , and 3 .

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