Question 588367
{{{(x+1)/(x-2)+(x-3)/(x-1)<0}}}
Maybe we could get a common denominator:
{{{(x+1)(x-1)/((x-2)(x-1))+(x-3)(x-2)/((x-2)(x-1))<0}}}
And we could do those multiplications in the numerator:
{{{(x^2-1)/((x-2)(x-1))+(x^2-5x+6)/((x-2)(x-1))<0}}}
And we could add the two fractions:
{{{(2x^2-5x+5)/((x-2)(x-1))<0}}}
The numerator is always positive.
We can see that {{{2x^2-5x+5=0}}} has no real roots because the discriminant is negative:
{{{(-5)^2-4*2*5=25-40=-15<0}}}
{{{2x^2-5x+5}}} is positive for x=0 (it's 5), and everywhere else.
For {{{(2x^2-5x+5)/((x-2)(x-1))}}} to be negative, the denominator has to be negative.
Each factor has a zero (x=1 and x=2), is negative for lesser x values and positive for greater values.
The product will be negative for x such that {{{1<x<2}}}.
For x=1 and x=2, the denominator is zero and the functions do not exist.
For x<1 and x>2 the function is positive.
So the solution is {{{highlight(1<x<2)}}}