Question

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

please answer this ASAP​

Answer 1

(3x+4y)^2=169

(3x-4y)^2+48xy=169

3x-4y=√169-48=√121=11

Answer 2

Answer:

±11

Step-by-step explanation:

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