8y+5>(3y-2)/2 >2y — step-by-step solution

1) We have the following 2 conditions:

[1]:

8 y + 5 > \frac{3 y - 2}{2}

[2]:

\frac{3 y - 2}{2} > 2 y

2) Solving for y:

8 y + 5 > \frac{3 y - 2}{2}

[1]

3) Let's sum the fractions:

8 y + 5 - \frac{3 y - 2}{2} > 0

ⅰ

8 y - \frac{3 y - 2}{2}

\frac{2 \cdot 8 y - (3 y - 2)}{2}

\frac{2 \cdot 8 y - (3 y - 2)}{2} + 5 > 0

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 8

16

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

- (3 y - 2)

- 3 y + 2

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

\frac{16 y - 3 y + 2}{2} + 5

\frac{(16 y - 3 y + 2) + 5 \cdot 2}{2}

\frac{(16 y - 3 y + 2) + 5 \cdot 2}{2} > 0

ⓘ Like terms are grouped by adding their coefficients:

16 y - 3 y

13 y

■Multiplications are done before additions and subtractions:

5 \cdot 2

10

\frac{13 y + 2 + 10}{2} > 0

ⓘ The numerator and the denominator are computed before the division:

2 + 10

12

\frac{13 y + 12}{2} > 0

⇥The expression

13 y + 12

is zero at:

⇥

y = - \frac{12}{13}

⇥Its coefficient of

y

13

, being positive, it is negative before and positive after

- \frac{12}{13}

Solution found [1]:

y > - \frac{12}{13}

4) Let's solve for y:

\frac{3 y - 2}{2} > 2 y

[2] where

y > - \frac{12}{13}

5) Let's sum the fractions:

\frac{3 y - 2}{2} - 2 y > 0

ⅰ

\frac{3 y - 2}{2} - 2 y

\frac{(3 y - 2) + 2 \cdot (- 2 y)}{2}

\frac{(3 y - 2) + 2 \cdot (- 2 y)}{2} > 0

ⓘ Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-2)

-4

\frac{3 y - 2 - 4 y}{2} > 0

ⓘ Like terms are grouped by adding their coefficients:

3 y - 4 y

- y

\frac{- y - 2}{2} > 0

ⓘ The minus sign is factored out, changing the sign of each term:

- y - 2

- (y + 2)

- \frac{y + 2}{2} > 0

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

- (y + 2)

- y - 2

⇥

- y - 2

⇥The expression

- y - 2

is zero at:

⇥

y = -2

⇥Its coefficient of

y

-1

, being negative, it is positive before and negative after

-2

No value of

y

satisfies both

y > - \frac{12}{13}

and

y < -2

▶Solution:

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