Solve the following system of equation by elimination method.
The given equations are
+ 
= 
… (1)
+ 
= 0 … (2)
Let a = 
and b =
.
⇒ 2a + 
b = 
… (3)
⇒ 3a + 2b = 0 … (4)
Now, (3) × 3 – (4) × 2
⇒ 6a + 2b – (6a + 4b) = 1/2 – 0
⇒ 6a + 2b – 6a – 4b = 1/2
⇒ – 2b = 1/2
⇒ b = 
Substituting b = 
in (4),
⇒ 3a + 2(
) = 0
⇒ 3a = 1/2
⇒ a = 
When a =
, 
=
. Thus, x = 6.
When b =
, 
=
. Thus, y = – 4
∴ (6, – 4) is the solution to the given system.
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