# SOLVING INEQUALITIES BY CLEARING THE FRACTIONS

To solve for the variable indicated in the question, we have to isolate the variable term in one side. For that, we will use inverse operations.

• Inverse operation for addition is subtraction.
• Inverse operation for subtraction is addition.
• Inverse operation for multiplication is division.
• Inverse operation of division is multiplication

To get rid of the fraction, we have to multiply it by the multiplicative inverse.

Problem 1 :

(-3/4)m - (1/8) ≤ (-1/4)

Solution :

(-3/4)m + (-1/8) ≤ (-1/4)

(-3/4)m + (-1/8) + 1/8 ≤ (-1/4) + 1/8

(-3/4)m ≤ (-1/8)

Since we multiply both sides by (-4/3), change inequality ≥ into ≤

(-3/4m) (-4/3) ≤ (-1/8) (-4/3)

m ≥ 1/6

Problem 2 :

(7/13)x  - 1 > 1/2

Solution :

(7/13)x  - 1 + 1 > (1/2) + 1

(7/13)x > 3/2

Multiply both sides by 13/7

(7/13)x (13/7) > (3/2) (13/7)

x > 39/14

Converting improper fraction into mixed fraction, we get

x > 2 11/14

Problem 3 :

(4/5) ≥ (2/3) - (2/7x)

Solution :

(4/5) ≥ (-2/7x) + (2/3)

Subtract 2/3 from both sides.

(4/5) - 2/3 ≥ (-2/7x) + 2/3 - 2/3

2/15 ≥ (-2/7)x

Since we multiply both sides by (-7/2), change inequality ≥ into ≤

(2/15) (-7/2≥ (-2/7)x (-7/2)

-7/15 ≤ x

Problem 4 :

(8/15x) – (17/30) < 7/10

Solution :

(8/15x) – (17/30) + 17/30 < (7/10) + 17/30

8/15x < 38/30

Multiply both sides by 15/8

(8/15)x (15/8) < (38/30) (15/8)

x < 19/8

Problem 5 :

(-4/11)z - 1 > (-8/11)

Solution :

(-4/11)z – 1 + 1 > (-8/11) + 1

(-4/11)z > 3/11

Since we multiply both sides by (-11/4), change inequality > into <

(-4/11)z (-11/4) > (3/11) (-11/4)

z < -3/4

Problem 6 :

(1/5k) + 14 ≤ 2/9

Solution :

Subtract 14 from both sides

(1/5k) + 14 – 14 ≤ 2/9 – 14

1/5k ≤ -124/9

Multiply both sides by 5

(1/5)k (5) ≤ (-124/9) (5)

k ≤ -620/9

Converting improper fraction into mixed fraction, we get

k ≤ -68  8/9

Problem 7 :

-31/4 < -13 + (7/8f)

Solution :

-31/4 < -13 + (7/8f)

-31/4 +13 < -13 + 13 + (7/8)f

21/4 < 7/8f

Multiply both sides by 8/7

(21/4) (8/7) < (7/8f) (8/7)

6 < f

Problem 8 :

(1/7r) + (53/56) > 6/7

Solution :

Subtract 53/56 from both sides

(1/7)r + (53/56) – (53/56) > (6/7) – (53/56)

1/7r > -5/56

Multiply both sides by 7

(1/7)r (7) > (-5/56) (7)

r > -5/8

Problem 9 :

(5/6n) – (1/5) < -8/15

Solution :

(5/6)n – (1/5) + (1/5) < (-8/15) + (1/5)

5/6n < -5/15

Multiply both sides by 6/5

(5/6n) (6/5) < (-5/15) (6/5)

n < -2/5

Problem 10 :

(1/3) + (1/13d) ≥ 17/39

Solution :

Subtract 1/3 from both sides

(1/3) - (1/3) + (1/13)d ≥ (17/39) - (1/3)

1/13d ≥ 14/39

Multiply both sides by 13

(1/13d) (13) ≥ (14/39) (13)

d ≥ 14/13

d ≥ 1 1/13

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