Question

if 4x^2+9y^2=16 and xy=1 find the value of 2x-3y

Answer 1

Solution :

Given,

xy = 1 ----( 1 )

4x² + 9y² = 16

=> (2x)² + (3y)²

-2*2x*3y=16-2*2x*3y

=> ( 2x - 3y )² = 16 - 12xy

=> ( 2x - 3y )² = 16 - 12

[ from (1) ]

=> ( 2x - 3y )² = 4

=> 2x - 3y = ± √4

= ± 2

•••••

Answer 2

To find : 2x - 3y



Let the algebraic value of 2x - 3y be a,

= > 2x - 3y = a


Square on both sides,

= > ( 2x - 3y )^2 = a^2



==================
From the properties of factorization, we know

( a - b )^2 = a^2 + b^2 - 2ab
=================



= > ( 2x )^2 + ( 3y )^2 - 2( 2x × 3y ) = a^2


= > 4x^2 + 9y^2 - 12xy = a^2



=================
In the question,

4x^2 + 9y^2 = 16

xy = 1
================



= > 16 - 12( 1 ) = a^2

= > 16 - 12 = a^2

= > 4 = a^2

= > ±2 = a

= > ± 2 = 2x - 3y



Hence the numeric value of 2x - 3y is ± 2 .
$\:$