\[x\]
such that\[5 \lt 3x-1 \lt 14\]
\[0 \lt 2x-2 \lt 10\]
From the first of these,
\[5+1 \lt 3x \lt 14+1 \rightarrow \frac{5+1}{3} \lt x \lt \frac{14+1}{3}\]
&Hence
\[2 \lt x \lt 5\]
From the second,
\[0+2 \lt 2x \lt 10+2 \rightarrow \frac{0+2}{2} \lt x \lt \frac{10+2}{2}\]
&Hence
\[1 \lt x \lt 6\]
\[x\]
must satisfy both inequalities. From the first \[2 \lt x\]
and from the second \[1 \lt x\]
. This last requirement is redundant since if \[2 \lt x\]
, \[1 \lt x\]
is satisfied. Also, the are some numbers eg 1.5, that satisfy the second inequality but not the first, so do not satisfy both.Similarly
\[x \lt 5\]
and \[x \lt 6\]
. The second of these is again redundant. The set of values of \[x\]
that satisfy both inequalities is \[2 \lt x \lt 5\]
.