$5x+2<3x+8$ and $(x+2)/(x-1)<4$

Question

The set of values of $x$ which satisfy the inequations $5x+2<3x+8$ and $(x+2)/(x-1)<4$ is?

I have done it and my answer is $(2,3)$ but it's wrong. The answer is given $(-\infty,1)$ union $(2,3)$.

Answer 1

First inequality gives $x < 3$ the second $\frac{x+2}{x-1} - 4 < 0 \rightarrow \frac{-3(x-2)}{x-1} < 0$ thus from the second inequality we have that $x > 2$ and $x < 1$. Finally we have that $x < 1$ and $2 < x < 3$

Answer 2

From the first

$$5x+2<3x+8 \iff2x<6\iff x<3$$

Now from the second

$$\frac{x+2}{x-1}<4$$

we set $x\neq1$ and distiguish two cases with

• $x-1 >0 \implies x+2<4(x-1)$
• $x-1 <0 \implies x+2>4(x-1)$

from here you can easily conclude.