Solve the equation and check the result : 3(2k + 1) – 2(k – 5)

1.	Solve the equation and check the result : 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16
Answer» Given 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16 ⇒ 6k + 3-2k + 10-25 + 10k = 16 (Distributive property) ⇒ 14k – 12 = 16 ⇒ 14k – 12 + 12 – 16 + 12 (Add 12 on both sides) ⇒ 14k = 28 ⇒ 14k/14 = 28/14 (Divide by 14 on both sides) ⇒ k = 2 Check: Substitute k = 2 in 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16 LHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 3[2 × 2 + 1] – 2[2 – 5] – 5[5 – 2 × 2] = 3[4 + 1] – 2(- 3) – 5(5 – 4) = 15 + 6 – 5 – 16 = RHS Hence verified.

Solve the equation and check the result : 3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16

Answer»

Given 3(2k + 1) – 2(k T 5) – 5(5 – 2k) = 16

⇒ 6k + 3-2k + 10-25 + 10k = 16 (Distributive property)

⇒ 14k – 12 = 16

⇒ 14k – 12 + 12 – 16 + 12 (Add 12 on both sides)

⇒ 14k = 28

⇒ 14k/14 = 28/14 (Divide by 14 on both sides)

⇒ k = 2

Check:

Substitute k = 2 in

3(2k + 1) – 2(k – 5) – 5(5 – 2k) = 16

LHS = 3(2k + 1) – 2(k – 5) – 5(5 – 2k)

= 3[2 × 2 + 1] – 2[2 – 5] – 5[5 – 2 × 2]

= 3[4 + 1] – 2(- 3) – 5(5 – 4)

= 15 + 6 – 5 – 16 = RHS

Hence verified.