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Question

Find the values of x and y if :   

52+x+1y4=2 

62+x3y4=1

Here, x -2 and y 4.


  1. x = 1, y = 7

  2. x = -2,  y = 2

  3. x = 0, y = 8

  4. x = 7, y = -8 


Solution

The correct option is A

x = 1, y = 7


Let p=12+x and  q=1y4

Thus, 5p + q = 2    .... (i)

6p - 3q = 1             .... (ii)

On multiplying equation (i) by 3, we get

15p + 3q = 6          .... (iii)

On adding equations (iii) and (ii), we get

21p = 7 

p=13 

On substituting the value of p in equation (i), we get

5×13+q=2

q=253

q=13

Now, p=12+x  

13=12+x 

x=1

Similarly q=1y4

13=1y4 

y=7

x = 1 and y = 7

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