5/(2t-1)<2 — step-by-step solution

1) Let's solve in

ℝ

\frac{5}{2 t - 1} < 2

2) Let's add fractions:

\frac{5}{2 t - 1} - 2 < 0

ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{5}{2 t - 1} - 2

\frac{5 - 2 \cdot (2 t - 1)}{2 t - 1}

\frac{5 - 2 \cdot (2 t - 1)}{2 t - 1} < 0

We study the sign of the numerator

Let's study the sign of

5 - 2 \cdot (2 t - 1)

by solving in

ℝ

5 - 2 \cdot (2 t - 1) < 0

ⓘ Expand by multiplying each term inside the parentheses:

- 2 \cdot (2 t - 1)

- (4 t - 2)

5 - (4 t - 2) < 0

ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (4 t - 2)

- 4 t + 2

5 - 4 t + 2 < 0

ⓘ Start from

5

and move forward

2

on the number line:

5 + 2

7

7 - 4 t < 0

ⓘ We add

4 t

to both sides

7 < 4 t

ⓘ Both sides are divided by

4

Candidates, pending the domain check:

t > \frac{7}{4}

⇥So the expression

5 - 2 \cdot (2 t - 1)

is negative on

(\frac{7}{4}, +∞)

and positive or zero otherwise

We study the sign of the denominator, which must not be zero

⇥The expression

2 t - 1

is zero at:

⇥

t = \frac{1}{2}

⇥Its coefficient of

t

2

, being positive, it is negative before and positive after

\frac{1}{2}

(forbidden value)

3) Let's draw up the sign table:

Candidates, pending the domain check:

t \in (-∞, \frac{1}{2}) \cup (\frac{7}{4}, +∞)

4) Let's find the domain of

\frac{5}{2 t - 1} < 2

For the expression

\frac{5}{2 t - 1}

to exist, you must check:

2 t - 1 \neq 0

ⓘ We add

1

to both sides

2 t \neq 1

ⓘ Both sides are divided by

2

Domain constraints:

t \neq \frac{1}{2}

▶Solution:

t \in (-∞, \frac{1}{2}) \cup (\frac{7}{4}, +∞)

5) The numeric value of:

t \in (-∞, \frac{1}{2}) \cup (\frac{7}{4}, +∞)

is:

▷

t \in (-∞, 0.5) \cup (1.75, +∞)