Question

Solve these (please help )​

Answer 1

$\Large{\underline{\underline{\mathfrak{\green{\bf{Solution}}}}}}$

$\Large{\underline{\sf{\bf{Given}}}}$

• 2x + 3y = 0 ........(1)
• (a-b) x + (a+b)y - (3a+b-2) = 0

Or,

• (a-b) x + (a+b)y = (3a+b-2) .....(2)

$\Large{\underline{\sf{\bf{Find}}}}$

• Value of x

$\Large{\underline{\underline{\mathfrak{\green{\bf{Explanation}}}}}}$

Firstly,

$\:\:\:\small\sf{\:Multiply\:by\:(a-b)\:in\:(1)\:and\:2\:in\:(2)}$

• 2(a-b)x + 3(a-b)y = 0
• 2(a-b)x + 6(a+b)y = 2(3a+b-2)

_______________Subtract this,

$\mapsto\sf{\:3(a-b)-6(a+b)y\:=\:-(6a+2b-4)}$

$\mapsto\sf{\:(3a-3b-6a-6b)y\:=\:-(6a+2b-4)}$

$\mapsto\sf{\:-(3a+9b)y\:=\:-(6a+2b-4)}$

$\mapsto\sf{\bf{\:y\:=\:\dfrac{(6a+2b-4)}{(3a+9b)}}}$

_____________________

$\:\:\:\:\small\sf{\:keep\:value\:in\:equ(1)}$

$\mapsto\sf{\:2x+3\times \left(\dfrac{(6a+2b-4)}{(3a+9b)}\right)\:=\:0}$

$\mapsto\sf{\:2x\:=\:\dfrac{-3(6a+2b-4)}{3(a+3b)}}$

$\mapsto\sf{\bf{\:x\:=\:\dfrac{-(6a+2b-4)}{2(a+3b)}}}$

______________________

$\bold{\underbrace{Thus}}$

$\mapsto\sf{\:Value\:of\:x\:=\:\dfrac{-(6a+2b-4)}{2(a+3b}}$

$\mapsto\sf{\:Value\:of\:y\:=\:\dfrac{(6a+2b-4)}{(3a+9b)}}$

_____________________

$\Large{\underline{\underline{\mathfrak{\green{\bf{Answer Verification}}}}}}$

$\:\:\:\:\small\sf{\:Keep\:Value\:of\:x\:and\:y\:in\:equ(1)}$

Take L.H.S.,

$\mapsto\sf{\:2\times \dfrac{-(6a+2b-4)}{2(a+3b)}+3\times \dfrac{(6a+2b-4)}{3(a+3b)}}$

$\mapsto\sf{\:\dfrac{(-6a-2b+4)}{a+3b)}\:+\:\dfrac{(6a+2b-4)}{a+3b)}}$

$\mapsto\sf{\:\dfrac{(-6a-2b+4+6a+2b-4)}{a+3b)}}$

$\mapsto\sf{\:\dfrac{0}{(a+3b)}}$

$\mapsto\sf{\:0}$

=R.H.S.

That's proved

______________________

Answer 2

Answer:

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