3-2x<= 4x+1<= 2x+7 — step-by-step solution

1) We have the following 2 conditions:

[1]:

3 - 2 x \leq 4 x + 1

[2]:

4 x + 1 \leq 2 x + 7

2) Let's solve for x:

3 - 2 x \leq 4 x + 1

[1]

3 - 2 x \leq 4 x + 1

ⓘ We gather the terms in

x

on the right and the other terms on the left

3 - 1 \leq 4 x + 2 x

ⓘ Start from

3

and move back

1

on the number line:

3 - 1

2

■Like terms are grouped by adding their coefficients:

4 x + 2 x

6 x

2 \leq 6 x

ⓘ Both sides are divided by

6

x \geq \frac{2}{6}

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

2

, their greatest common divisor:

\frac{2}{6}

\frac{1 \cdot 2}{3 \cdot 2}

\frac{1}{3}

Solution found [1]:

x \geq \frac{1}{3}

3) Solving for x:

4 x + 1 \leq 2 x + 7

[2] using

x \geq \frac{1}{3}

4 x + 1 \leq 2 x + 7

ⓘ We subtract

1

from both sides■Start from

7

and move back

1

on the number line:

7 - 1

6

4 x \leq 2 x + 6

ⓘ We subtract

2 x

from both sides

4 x - 2 x \leq 6

ⓘ Like terms are grouped by adding their coefficients:

4 x - 2 x

2 x

2 x \leq 6

ⓘ Both sides are divided by

2

x \leq \frac{6}{2}

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

2

, their greatest common divisor:

\frac{6}{2}

3

Solution found [2]:

x \leq 3

Intersecting

x \geq \frac{1}{3}

and

x \leq 3

gives

x \in [\frac{1}{3}, 3]

▶Solution:

x \in [\frac{1}{3}, 3]

4) The numeric value of:

x \in [\frac{1}{3}, 3]

equals:

▷

x \in [\approx 0.3333, 3]