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  • 23-04-2022
  • Mathematics
contestada

solve for x: 6 < 4x + 8 < 14

solve for x 6 lt 4x 8 lt 14 class=

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Аноним Аноним
  • 23-04-2022

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

Let's solve ~

[tex]\qquad \tt \dashrightarrow \:6 < 4x + 8 < 14[/tex]

[tex]\qquad \tt \dashrightarrow \:6 - 8 < 4x + 8 - 8 < 14 - 8[/tex]

[tex]\qquad \tt \dashrightarrow \: - 2 < 4x < 6[/tex]

[tex]\qquad \tt \dashrightarrow \: \dfrac{ - 2}{4} < \dfrac{4x}{4} < \dfrac{6}{4} [/tex]

[tex]\qquad \tt \dashrightarrow \: \dfrac{ - 1}{2} < {x}{} < \dfrac{3}{2} [/tex]

hope it helps ~

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