Question

If 3x + 4y = 13 , xy =1 find 3x -4y

Answer 1

Given,

3x + 4y = 13

xy= 1

To find,

3x-4y = ?

A/q,

=>(3x +4y)² = 9x² + 16y² + 24xy

=>(13)² = 9x² + 16y² + 24xy

=> 169 = 9x² + 16y² + 24(1) putting the value of xy

=> 169 = 9x² + 16y² + 24

=> 169-24 = 9x²+16y²

=> 145 = 9x² +16y²

Now,

=> (3x-4y)² = 9x² +16y²-24xy

=> (3x-4y)² = 145 -24(1)

=> (3x-4y)² = 121

=> 3x - 4y= √121

=> 3x-4y = 11

______________________

Answer 2

= 3x + 4y = 13 , xy=1 , 3x -4y =?

$(a+b)^{2} -(a-b)^2 = 4ab (Formula)\\(3x+4y)^2 - (3x-4y)^2= 4*3x*4y\\(13)^2 - (3x-4y)^2 =48*1\\169 - (3x-4y)^2 =48\\169-48= (3x-4y)^2\\121 = (3x-4y)^2\\11= 3x - 4y$