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15(y − 4) − 2(y − 9) + 5(y + 6) = 0

⇒ 15y − 60 − 2y + 18 + 5y + 30 = 0 (Opening the brackets)

⇒ 18y − 12 = 0

⇒ 18y = 12

⇒

L.C.M. of the denominators, 2, 4, and 6, is 12.

Multiplying both sides by 12, we obtain

6n − 9n + 10n = 252

⇒ 7n = 252

L.C.M. of the denominators, 2, 3, and 6, is 6.

Multiplying both sides by 6, we obtain

6x + 42 − 16x = 17 − 15x

⇒ 6x − 16x + 15x = 17 − 42

⇒ 5x = −25

L.C.M. of the denominators, 3 and 4, is 12.

Multiplying both sides by 12, we obtain

3(3t − 2) − 4(2t + 3) = 8 − 12t

⇒ 9t − 6 − 8t − 12 = 8 − 12t (Opening the brackets)

⇒ 9t − 8t + 12t = 8 + 6 + 12

⇒ 13t = 26

3(t − 3) = 5(2t + 1)

⇒ 3t − 9 = 10t + 5 (Opening the brackets)

⇒ −9 − 5 = 10t − 3t

⇒ −14 = 7t

3(5z − 7) − 2(9z − 11) = 4(8z − 13)−17

⇒ 15z − 21 − 18z + 22 = 32z − 52 − 17 (Opening the brackets)

⇒ −3z + 1 = 32z − 69

⇒ −3z − 32z = −69 − 1

⇒ −35z = −70

⇒