Solve the equation. Also, verify the result.(x/2) + (3/2) = (2x/5) –

1.	Solve the equation. Also, verify the result.(x/2) + (3/2) = (2x/5) – 1
Answer» Given (x/2) + (3/2) = (2x/5) – 1 Transposing (2x/5) to LHS and (3/2) to RHS, then we get (x/2) – (2x/5) = – 1 – (3/2) (5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10] x/10 = -5/2 Multiplying both sides by 10 we get, x/10 × 10 = (-5/2) × 10 x = (-50/2) x = -25 Verification: Substituting x = -25 in given equation we get (-25/2) + (3/2) = (-50/5) – 1 (-25 + 3)/2 = -10 – 1 (-22/2) = -11 -11 = -11 Thus LHS = RHS Hence, verified.

Solve the equation. Also, verify the result.(x/2) + (3/2) = (2x/5) – 1

Answer»

Given (x/2) + (3/2) = (2x/5) – 1

Transposing (2x/5) to LHS and (3/2) to RHS, then we get

(x/2) – (2x/5) = – 1 – (3/2)

(5x -4x)/10 = (-2 – 3)/2 [LCM of 5 and 2 is 10]

x/10 = -5/2

Multiplying both sides by 10 we get,

x/10 × 10 = (-5/2) × 10

x = (-50/2)

x = -25

Verification:

Substituting x = -25 in given equation we get

(-25/2) + (3/2) = (-50/5) – 1

(-25 + 3)/2 = -10 – 1

(-22/2) = -11

-11 = -11

Thus LHS = RHS

Hence, verified.