Question

solve.
\\(\frac{x + 3}{x - 1} < 1\\)

the solution set is \\(\square\\).
(type your answer in interval notation.)

Explanation:

Step1: Subtract 1 from both sides

To solve the inequality \(\frac{x + 3}{x - 1}<1\), we first subtract 1 from both sides to get a single fraction on the left - hand side.
\(\frac{x + 3}{x - 1}-1<0\)
We rewrite 1 as \(\frac{x - 1}{x - 1}\) (since \(x
eq1\), the denominator \(x - 1\) is non - zero), then:
\(\frac{x + 3-(x - 1)}{x - 1}<0\)

Step2: Simplify the numerator

Simplify the numerator \(x + 3-(x - 1)\):
\(x+3 - x + 1=\frac{4}{x - 1}\)
So the inequality becomes \(\frac{4}{x - 1}<0\)

Step3: Analyze the sign of the fraction

For a fraction \(\frac{a}{b}\) (where \(a = 4>0\) in our case), \(\frac{a}{b}<0\) when \(b<0\) (because a positive number divided by a negative number gives a negative number).
We have the denominator \(x - 1\), and we want \(x - 1<0\)
Solving \(x - 1<0\) gives \(x<1\)

Answer:

\((-\infty,1)\)