- Thread starter geekay
- Start date

if I open the parentheses on the RHS, it then becomes |-1*(a-2b)|=|-a+2b|, would it not?

Guess this is wrong way of thinking. So should one instead use |-1*(a-2b)|=|-1|*|a-2b|=|a-2b| ?

Perhaps my problem is this: don’t know how to get rid of the absolute value signs.

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- Nov 12, 2017

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Recall the piecewise definition of absolute value: if x>=0, then |x| = x; if x<0, then |x| = -x.If 0<a<2b applies, then

|a-2b| = 2b-a

Could someone please explain me why the above is true?

Sorry if this is posted in the wrong section.

Much appreciated.

Which case applies here?

Good so far?

\(\displaystyle \text {Let } d \text { be any real number.}\)

\(\displaystyle \therefore d < 0, d = 0, \text { or else } d > 0.\)

\(\displaystyle \text {CASE I: } d < 0 \implies \\

|\ d \ | = - d > 0 \implies |\ - d \ | = - d \implies |\ d \ | = |\ - d \ |.\)

\(\displaystyle \text {CASE II: } d = 0 \implies \\

|\ d \ | = 0 = - 0 \implies |\ - d \ | = |\ - 0 \ | = |\ 0 \ | = 0 \implies |\ d \ | = |\ - d \ |.\)

\(\displaystyle \text {CASE III: } d > 0 \implies \\

|\ d \ | = d > 0 \implies - d < 0 \implies |\ - d \ | = - (-d) = d \implies |\ d \ | = |\ - d \ |.\)

\(\displaystyle \therefore \text {in all cases, } |\ d \ | = |\ - d \ |.\)

Any questions?

Now apply that theorem to your problem.