text: Thread: Rewrite as a double inequality?
  
  1. Rewrite as a double inequality?
  
  I stumbled upon this question: "Rewrite |5x+7| < 10 as double inequality".
  
  My attempt:
  
  5x+7 < 10 and (-5x+7)<10
  5x < 3 and -5x<17
  x <3/5 and x>17/ 5
  
  Is that the solution? The given options are :
  
  a. -10> 5x +7 >10
  b. -10< 5x+7<10
  c. -10 < 5x+7>10
  d. 0<5x+7<10
  
  (I know c and a are wrong because of the direction of the greater than sign. To me b. seems right)
  
  2. Re: Rewrite as a double inequality?
  
  Originally Posted by sakonpure6
  I stumbled upon this question: "Rewrite |5x+7| < 10 as double inequality".
  This depends upon this theorem: $|a|\le |b|\text{ if and only if }-|b|\le a\le |b|$.
  
  So in this case $-10<5x+7<10$.
  
  3. Re: Rewrite as a double inequality?
  
  So the value of b needs to always be positive? Does this work |5x+7| < -10 ? No? May i please get the name of this theorem. Thank you!
  
  4. Re: Rewrite as a double inequality?
  
  Originally Posted by sakonpure6
  So the value of b needs to always be positive? Does this work |5x+7| < -10 ? No? May i please get the name of this theorem.
  It is such a basic property of the metric that it has no name.