The set of values of x which satisfy 5x + 2

Question

(B) Step $1$: Solve the first inequality $5x + 2 < 3x + 8$.$5x - 3x < 8 - 2$$2x < 6$$x < 3$.Step $2$: Solve the second inequality $\frac{x + 2}{x - 1}

Vedclass · Accepted Answer

$(-\infty, 1) \cup (2, 3)$

Vedclass · Answer

(B) Step $1$: Solve the first inequality $5x + 2 $5x - 3x $2x $x Step $2$: Solve the second inequality $\frac{x + 2}{x - 1} Subtract $4$ from both sides: $\frac{x + 2}{x - 1} - 4 $\frac{x + 2 - 4(x - 1)}{x - 1} $\frac{x + 2 - 4x + 4}{x - 1} $\frac{-3x + 6}{x - 1} Multiply by $-1$ and reverse the inequality sign: $\frac{3x - 6}{x - 1} > 0$.$\frac{3(x - 2)}{x - 1} > 0$.The critical points are $x = 1$ and $x = 2$. Testing intervals $(-\infty, 1)$,$(1, 2)$,and $(2, \infty)$,the expression is positive in $(-\infty, 1) \cup (2, \infty)$.Step $3$: Find the intersection of $x The intersection is $(-\infty, 1) \cup (2, 3)$.

The set of values of $x$ which satisfy $5x + 2 < 3x + 8$ and $\frac{x + 2}{x - 1} < 4$ is:

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