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4/x=5y=7

3/x+4y=5

By elimination method;

4/x+5y=7 -- (1) and 3/x+4y=5 -- (2)

Now, (1/x=p and 1y=q)

4p+5q=7 (1) and 3p+4q=5 (2)

Multiplying (1) with 4 and (2) with 5, we get

16p+20q=28

15p+20q=25 (Subtracting)

p=3

And, Now substituting found value of p in (1)

4(3)+5q=7

12+5q=7

5q=-5

q=-5/5=-1

Now to find values of x and y,

1/x=3

x=3 &

y=-1.

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Let, 1/x= 'a' & 1/y= 'b', 

Thrfor, eq will be, 

4a - 5b=7------(1) 

3a - 4b=5------(2) 

Now by sbstitutn method, 

from eq (1) 

  a= (7+ 5b) /4-----(3)

sbstitutng this value in eq (2)  

3 [ (7+5b) /4 ] - 4b=5  

=> (21 + 15b) /4 - 4b=5

=> (21 + 15b - 16b) / 4=5                   (taking the LCM)

=>  21-b=20

 -b = -1 => b=1

Now sbstitutng b=1 in eq (3)  we get,

 a= (7 + 5x1) /4 

=> a= 12/4  => a=3

Thrfor,

  1/x= 3   => x=1/3    

and,  1/y=1  => y=1

Hence,  x=1/3 and y=1  

                                             

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