Question 875257
That kind of problem always has infinite solutions.
{{{3x-1}}} and {{{x-3}}} are linear expressions in {{{x}}} .
As {{{x}}} increases the value of those expressions increases.
As {{{x}}} decreases the value of those expressions decreases.
 
{{{3x-1<24}}}
If you find that for some {{{x}}} , say {{{x=8}}} ,
the value of {{{3x-1}}} is less than {{{24}}} ,
then any other smaller x will also be a solution.
{{{3*8-1=24-1=23<24}}} so {{{x=8}}} is a solution,
but so is any number smaller than {{{8}}},
such as {{{-1000}}}, {{{-3/5}}} , {{{"0"}}} , {{{1}}}, {{{sqrt(2)}}} , {{{pi}}} .
There are infinite solutions, so we cannot list them all.
We have to give a formula (an inequality) that shows all the solutions
{{{3x-1<24}}} --> {{{3x-1+1<24+1}}} --> {{{3x<25}}} --> {{{3x/3<25/3}}} --> {{{highlight(x<35/3)}}}
Since {{{25/3=8&1/3}}} , that can also be written as {{{highlight(x<8&1/3)}}}
 
{{{x-3<8}}} --> {{{x-3+3<8+3}}} --> {{{highlight(x<11)}}}
Some of the infinite solutions are {{{10.999}}} , {{{1}}}, {{{-1/2}}} ,
and many other numbers that are less than 11.