Question

find the range of value of x which satisfy 5x+2<3x+8 and x+2÷x-1<4​

Answer 1

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Answer 2

Consider,

$\longrightarrow 5x+2<3x+8$

$\longrightarrow 2x<6$

$\longrightarrow x<3$

$\Longrightarrow x\in(3,\ \infty)\quad\quad\dots(1)$

Consider,

$\longrightarrow\dfrac{x+2}{x-1}<4$

$\longrightarrow\dfrac{x+2}{x-1}-4<0$

$\longrightarrow\dfrac{x+2-4(x-1)}{x-1}<0$

$\longrightarrow\dfrac{-3x+6}{x-1}<0$

$\longrightarrow\dfrac{x-2}{x-1}>0$

$\Longrightarrow x\in(-\infty,\ 1)\cup(2,\ \infty)\quad\quad\dots(2)$

Taking (1) and (2) in common we get,

$\longrightarrow x\in(3,\ \infty)\cap[(-\infty,\ 1)\cup(2,\ \infty)]$

$\longrightarrow\underline{\underline{x\in(3,\ \infty)}}$

This is the range of values of $x.$