- #1

chwala

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- Homework Statement
- See attached

- Relevant Equations
- modulus

This is the question * consider the highlighted question only *with its solution shown

My approach is as follows (alternative method),

##|\frac {5}{2x-3}| ##< ## 1##

Let, ##\frac {5}{2x-3} ##⋅##\frac {5}{2x-3}##=##1##

→##x^2-3x-4=0##

##(x+1)(x-4)=0##

it follows that the critical values are ##x_1=-1## and ##x_2=4##, which will give the correct solution once we confirm/establish the valid region in the given inequality.

Is their another method on this type of questions? ... regards

**(from textbook)**;My approach is as follows (alternative method),

##|\frac {5}{2x-3}| ##< ## 1##

Let, ##\frac {5}{2x-3} ##⋅##\frac {5}{2x-3}##=##1##

→##x^2-3x-4=0##

##(x+1)(x-4)=0##

it follows that the critical values are ##x_1=-1## and ##x_2=4##, which will give the correct solution once we confirm/establish the valid region in the given inequality.

Is their another method on this type of questions? ... regards