Question

Its a simultaneous equation ,if x+y=10 ,x^2+y^2=58, find x and y?

Answer 1

Hey dear !!!

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===>> In the example

x + y = 10 ...................(1)

x² + y² = 58 ..............(2)

x + y = 10

y = 10 - x ............(3)

Now, put the value of y in equation (2) we get,

x² + y² = 58

x² + ( 10 - x )² = 58

Now, here we will use the identity of [ ( a - b )² = a² - 2ab + b² ]

By applying the identity we get,

x² + 10² - 2 * 10 * x + x² = 58

x² + 100 - 20x + x² = 58

2x²- 20x + 100 - 58 = 0

2x² - 20x + 42 = 0

Now divide by 2 on both sides we get,

x² - 10x + 21 = 0

Now factorise it !!

x² - 7x - 3x + 21

x( x - 7 ) -3( x - 7 )

( x - 7 ) or ( x -3 )

So, possible value of x will be

x - 7 = 0

∴ x = 7

or

x -3 = 0

∴ x = 3

Now, putting the value of x in equation (3) we get,

If x = 7

y = 10 - x

y = 10 - 7

∴ y = 3

If x = 3

y = 10 - x

y = 10 - 3

∴ y = 7

So, ( x, y ) = (7,3) or (3,7) are the solution of the given equation !!

Hope you helped by me !

Any quarries regarding this feel free to ask !!

[ Be Brainly ]