Question

Solve the above simultaneous equation​

Answer 1

Step-by-step explanation:

Given :-

a/x - b/y = 0 and

ab^2/x + a^2b/y = (a^2+b^2)

To find:-

Solve the above simultaneous equations

Solution:-

Given equations are :

a/x - b/y = 0 ----------------------------(1) and

ab^2/x + a^2b/y = (a^2+b^2)-------(2)

put a/x = p and b/y = q then

above equations become

p-q = 0 -----------------------------------(3)

=>p=q -------------------------------------(4)

and

pb^2+qa^2=(a^2+b^2)----------------(5)

from (4) Substituting the value of p in (5)

=>qb^2+qa^2=(a^2+b^2)

=>q(a^2+b^2) = (a^2+b^2)

=>q= (a^2+b^2)/(a^2+b^2)

=>q=1

The value of q = 1

=>The value of p=1 ( since p=q)

now p=a/x

=>a/x =1

=>x=a

and

b/y=q

=>b/y=1

=>y=b

The value of x =a

The value of y =b

Answer:-

The solution for the given pair of equations is (a,b)

Check:-

equation -1:

LHS =a/x - b/y

=>a/a - b/b

=>1-1

=>0

=>RHS

LHS = RHS

equation-2:-

LHS:-

ab^2/x + a^2b/y

=>ab^2/a + a^2b/b

=>b^2+a^2

=>a^2 + b^2

=>RHS

LHS = RHS is true for x=a and y=b

The solution is (a,b) is true .