How do you solve $- 8 \leq 2 x - 4 < 6$ $-8<=2x-4<6$ ?

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Answer 1 · 2018-03-02T11:55:10

Using both inequalities. See below

From $- 8 \leq 2 x - 4$ $-8<=2x-4$ Sum 4 to both sides

$4 - 8 \leq 2 x - 4 + 4$ $4-8<=2x-4+4$

$- 4 \leq 2 x$ $-4<=2x$ and then $- 2 \leq x$ $-2<=x$ this is the same that $x \geq - 2$ $x>=-2$

Now, From $2 x - 4 < 6$ $2x-4<6$ Add 4 to both sides

$2 x - 4 + 4 < 6 + 4$ $2x-4+4<6+4$ and then

$2 x < 10$ $2x<10$ Which is $x < 5$ $x<5$

Collecting all results, we have $x$ $x$ must be $x \geq - 2$ $x>=-2$ and $x < 5$ $x<5$

This is the interval $[- 2, 5)$ $[-2, 5)$ The solution is all $x$ $x$ belonging to this interval $x \in [- 2, 5)$ $x in [-2,5)$ or using set notation

${x in RR// x>=-2 and x<5}$

Answer 2 · 2018-03-02T12:01:04

Solution : $-2<= x <5$ In interval notation: $x| [ -2 , 5)$

$-8 <= 2x-4 <6 or -8+4 <= 2x-4+4 <6+4$ or

$-4<= 2x <10 or -4/2<= (2x)/2 <10/2$ or

$-2<= x <5 :.$ Solution : $-2<= x <5$

In interval notation: $x| [ -2 , 5)$ [Ans]

How do you solve -8<=2x-4<6−8≤2x−4<6-8<=2x-4<6?