Help me solve this inequalities please...thanks

Question

1) 2/3 < (x+5)/-4 < 3/2

2)(6x+5)/2 >= to (4x-3)/5

Answer 1

1 )

2/3 < (x + 5 ) / - 4 < 3/2

Multiply each side by -4 and flip the symbol.

(2/3)(- 4) > [(x + 5)/- 4](- 4) > (3/2)(- 4)

(- 8/3) > (x + 5) > - 6

( -8/3) - 5 > x + 5 -5 > -6 -5

-23/3 > x > - 11

Solution set is {x | -11< x < -23/3 }.

2 )

 (6x + 5)/2 ≥ (4x - 3)/5

cross multiplication.

5(6x + 5) ≥ 2(4x - 3 )

30x + 25 ≥ 8x - 6

30x ≥ 8x - 31

30x  - 8x ≥ - 31

22x ≥ - 31

x ≥ - 31/22.

Solution set is {x | x ≥ -31/22 }.

Comments

Thank you :)

Answer 2

1).

2/3 < (x + 5)/- 4 <3/2

Multiply - 4 to each term of the inequality,flips the symbol.

(2/3)(- 4) > [(x + 5)/- 4](- 4) > (3/2)(- 4)

(- 8/3) > (x + 5) > - 6

Subtract 5 from each term.

(- 8/3) - 5 > (x + 5 -5) > - 6 - 5

Apply L.C.M rule : 1/a + 1/b = (b + a)/ab.

(- 8 - 15)/3 > x > - 11

- 23/3 > x > - 11.

Solution set is {x / x : - 23/3 > x > - 11}.

2).

(6x + 5)/2 ≥ (4x - 3)/5

Multiply 10 to each term.

[(6x + 5)/2]10 ≥ [(4x - 3)/5]10

5(6x + 5) ≥ 2(4x - 3)

Apply distributive property : a(b + c) = ab + ac

                                            a(b - c) = ab - ac.

30x + 25 ≥ 8x - 6

Subtract 5 from each term.

30x + 25 - 8x ≥ 8x - 6 - 8x

22x + 25 ≥ - 6

Subtract 5 from each term.

22x + 25 - 25  ≥ - 6 - 25

22x  ≥ - 31

Divide each term by 22.

x ≥ - 31/22.

Solution set is {x / x : x ≥ -31/22}.

Comments

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