Solve the equation t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t

1.	Solve the equation t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t -4). Check your result.
Answer» The above question can be written as,= t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12 Now, find the value of t by considering the above equation, Transposing -5, -5 to RHS and it becomes 5, 5 and 8t, -3t to LHS it becomes -8t, 3t = t – 2t + 10t – 8t + 3t = 6 + 12 + 5 + 5 = 14t – 10t = 28 = 4t = 28 Multiplying both side by (1/4), = 4t × (1/4) = 28 × (1/4) = t = 7 By substituting 7 in the place of t in given equation, we get LHS, = t – 2t – 5 – 5 + 10t = 7 – 2× (7) – 5 – 5 + 10(7) = 7 – 14 – 5 – 5 + 70 = 77 – 24 = 53 RHS, = 6 + 8t – 3t + 12 = 6 + 8 × (7) – 3 × (7) + 12 = 6 + 56 – 21 + 12 = 74 – 21 = 53 By comparing LHS and RHS = 15 = 15 ∴ LHS = RHS Hence, the result is verified.

Solve the equation t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t -4). Check your result.

Answer»

The above question can be written as,= t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12

Now, find the value of t by considering the above equation,

Transposing -5, -5 to RHS and it becomes 5, 5 and 8t, -3t to LHS it becomes -8t, 3t

= t – 2t + 10t – 8t + 3t = 6 + 12 + 5 + 5

= 14t – 10t = 28

= 4t = 28

Multiplying both side by (1/4),

= 4t × (1/4) = 28 × (1/4)

= t = 7

By substituting 7 in the place of t in given equation, we get

LHS,

= t – 2t – 5 – 5 + 10t

= 7 – 2× (7) – 5 – 5 + 10(7)

= 7 – 14 – 5 – 5 + 70

= 77 – 24

= 53

RHS,

= 6 + 8t – 3t + 12

= 6 + 8 × (7) – 3 × (7) + 12

= 6 + 56 – 21 + 12

= 74 – 21

= 53

By comparing LHS and RHS

= 15 = 15

∴ LHS = RHS

Hence, the result is verified.