Solve

Asked by  | 22nd Jul, 2010, 03:52: PM

Expert Answer:

Dear Student,
 
Here the given pair of equations are not linear. But we can change this to a linear pair using the substitution
 
Put  

    1          =   p         and
x   +   y      

      1        =  q
x   -   y      
 
the given equations become
 
10p + 2q = 4    .............(i)
 
15p - 5q = -2   .............(ii)
 
Let us solve these using elimination method
 
3(i) - 2(ii) gives
 
30p + 6q - 30p + 10q = 12 + 4
 
16q = 16
 
which means q = 1
 
Substituting this value of q in (i), we get
 
10p + 2(1) = 4
 
10p = 2
 
p = 1/5
 
So the solution to linear equations in p and q is
 
p = 1/5,   q = 1
 
But we want the answer in x and y
 
So
    1       = 1/5            which means x + y = 5   .......(iii)
x   +   y
 
and    
 
    1        = 1            which means x - y = 1   .........(iv)
 
x   -   y
 
We will solve (iii) and (iv) to get the final answer
 
Adding them, we get 2x = 6

So x = 3
 
Subtracting (iv) from (iii), we get 2y = 4

So y = 2
 
 
Hence, the final solution is   x = 3, y = 2
 
 
Regards,
 
Team Topper Learning

Answered by  | 22nd Jul, 2010, 06:58: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.