How do you solve for x in $3 x - 5 < x + 9 \leq 5 x + 13$ $3x-5 < x + 9 \le 5x + 13$ ?

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Answer 1 · 2014-10-31T15:36:56

By separating into two inequalities,

${(3x-5 < x+9),(x+9 le 5x+13):}$

Let us work on the first inequality.

$3x-5 < x+9$

by subtracting $x$ ,

$=> 2x-5 < 9$

by adding 5,

$=> 2x<14$

by dividing by $2$ ,

$=> x<7$

Let us work on the second inequality.

$x+9 le 5x+13$

by subtracting $x$ ,

$=> 9 le 4x+13$

by subtracting $13$ ,

$=> -4 le 4x$

by dividing by $4$ ,

$=> -1 le x$

By combining the two inequalities, we have

$-1 le x < 7$ ,

or in interval notation, the solution set is

$[-1,7)$ .

I hope that this was helpful.

How do you solve for x in 3x-5 < x + 9 \le 5x + 13 3x−5<x+9≤5x+133x-5 < x + 9 \le 5x + 13 ?