how to solve this ?? n16

[dev644]

 

[Deleted member 4993]

First, multiply all the segments by (3 * 2 =) 6 and get rid of the fractional terms.

 

[pka]

View attachment 11933

I assume that the question is \(\displaystyle x-3<\frac{2x-7}{3}\le\frac{3x+2}{2}\)
If it is then \(\displaystyle 6x-18<4x-14\le 9x+6\)
Then solve these two and look that the intersection.
\(\displaystyle 6x-18<4x-14\) and \(\displaystyle 4x-14\le 9x+6\)

 

[dev644]

I got that

I assume that the question is \(\displaystyle x-3<\frac{2x-7}{3}\le\frac{3x+2}{2}\)
If it is then \(\displaystyle 6x-18<4x-14\le 9x+6\)
Then solve these two and look that the intersection.
\(\displaystyle 6x-18<4x-14\) and \(\displaystyle 4x-14\le 9x+6\)

I got that but I am unable to solve it

 

[pka]

I got that I got that but I am unable to solve it

and
\(\displaystyle \begin{align*}6x-18&<4x-14 \\2x&<4\\x&<2 \end{align*}\) and \(\displaystyle \begin{align*}4x-14&\le 9x+6 \\-5x&\le 20\\x&\ge -4 \end{align*}\)

SEE HERE