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.