The equation is followed by the values of the variable. Decide whether

1.	The equation is followed by the values of the variable. Decide whether these values are the solution of that equation.2a + 4 = 0, a = 2, – 2, 1
Answer» 2a + 4 = 0 …..(i) Substituting a = 2 in L.H.S. of equation (i), L.H.S. = 2 (2) + 4 = 4 + 4 = 8 R.H.S. = 0 ∴ L.H.S. ≠ R.H.S. ∴ a = 2 is not the solution of the given equation. Substituting a = – 2 in L.H.S. of equation (i), L.H.S. = 2 (-2)+ 4 = -4 + 4 = 0 R.H.S. = 0 ∴ L.H.S. = R.H.S. ∴ a = – 2 is the solution of the given equation. Substituting a = 1 in L.H.S. of equation (i), L.H.S. = 2(1)+ 4 = 2 + 4 = 6 R.H.S. = 0 ∴ L.H.S. ≠ R.H.S. ∴ a = 1 is not the solution of the given equation.

The equation is followed by the values of the variable. Decide whether these values are the solution of that equation.2a + 4 = 0, a = 2, – 2, 1

Answer»

2a + 4 = 0 …..(i)

Substituting a = 2 in L.H.S. of equation (i),

L.H.S. = 2 (2) + 4

= 4 + 4

= 8

R.H.S. = 0

∴ L.H.S. ≠ R.H.S.

∴ a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i),

L.H.S. = 2 (-2)+ 4

= -4 + 4

= 0

R.H.S. = 0

∴ L.H.S. = R.H.S.

∴ a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),

L.H.S. = 2(1)+ 4

= 2 + 4

= 6

R.H.S. = 0

∴ L.H.S. ≠ R.H.S.

∴ a = 1 is not the solution of the given equation.