Question

x+y=5,x-y=1 then 8xy(x2+y2)

Answer 1

adding both the equations we get
x + y =5
x - y = 1
______
2x = 6
x = 3

replacing value of x in equation
x + y = 5
3 + y = 5
y = 2

putting the two values in the req eq
8xy(x2 + y2)
8(3)(2)[9 + 4]
48 * 13
= 624

Answer 2

x + y = 5 , x - y = 1 then 8xy(x² + y²) = 624

Given : x + y = 5 , x - y = 1

To find : The value of 8xy(x² + y²)

Solution :

Step 1 of 3 :

Find the value of 4xy

$\sf 4xy$

$\sf = {(x + y)}^{2} - {(x - y)}^{2}$

$\sf = {(5)}^{2} - {(1)}^{2}$

$\sf = 25 - 1$

$\sf = 24$

Step 2 of 3 :

Find the value of 2(x² + y²)

$\sf 2( {x}^{2} + {y}^{2} )$

$\sf = {(x + y)}^{2} + {(x - y)}^{2}$

$\sf = {(5)}^{2} + {(1)}^{2}$

$\sf = 25 + 1$

$\sf = 26$

Step 3 of 3 :

Find the value of 8xy(x² + y²)

8xy(x² + y²)

= 4xy × 2(x² + y²)

= 24 × 26

= 624

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