3<= (2x-3)/5 <7 — step-by-step solution

1) We have the following 2 conditions:

[1]:

3 \leq \frac{2 x - 3}{5}

[2]:

\frac{2 x - 3}{5} < 7

2) Let's solve for x:

3 \leq \frac{2 x - 3}{5}

[1]

\frac{2 x - 3}{5} \geq 3

2 x - 3 \geq 3 \cdot 5

ⅰ

3 \cdot 5

15

2 x - 3 \geq 15

ⓘ We add

3

to both sides

3) Let's calculate:

2 x \geq 15 + 3

ⓘ Start from

15

and move forward

3

on the number line:

15 + 3

18

2 x \geq 18

ⓘ Both sides are divided by

2

x \geq \frac{18}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{18}{2}

9

Solution obtained [1]:

x \geq 9

4) Let's solve for x:

\frac{2 x - 3}{5} < 7

[2] using

x \geq 9

\frac{2 x - 3}{5} < 7

2 x - 3 < 7 \cdot 5

ⅰ

7 \cdot 5

35

2 x - 3 < 35

ⓘ We add

3

to both sides

2 x < 35 + 3

ⓘ Start from

35

and move forward

3

on the number line:

35 + 3

38

2 x < 38

ⓘ Both sides are divided by

2

x < \frac{38}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{38}{2}

19

Solution found [2]:

x < 19

Intersecting

x \geq 9

and

x < 19

gives

x \in [9, 19)

▶Solution:

x \in [9, 19)