How do you solve $5 < 4 x + 3 \leq 11$ $5< 4x + 3\leq 11$ ?

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Answer 1 · 2017-09-21T16:41:55

What you do to the center to make it become only x, you do to all 3.

Given: $5 < 4 x + 3 \leq 11$ $5< 4x + 3 <= 11$

What you do to the center to make it become only x, you do to all 3.

Subtract 3:

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

Divide by 4:

$\frac{1}{2} < x \leq 2$ $1/2< x <= 2$

Solved

Answer 2 · 2017-09-21T17:35:51

$\frac{1}{2} < x \leq 2$ $1/2< x <=2$

Subtract 3 from both sides:

$2 < 4 x \leq 8$ $color(green)(2)<4x<=color(green)(8)$

Divide 4 from both sides:

$\frac{2}{4} < x \leq \frac{8}{4}$ $color(green)(2/4)< x <=color(green)(8/4)$

Simplify:

$\frac{1}{2} < x \leq 2$ $color(green)(1/2)< x <=color(green)(2)$

How do you solve 5<4x+3≤115< 4x + 3\leq 11?