निम्नलिखित समीकरण निकाय को हल करें-
`ax+by=c,bx+ay=1+c`

Explanation

दत्त समीकरण है `ax+by=c`…………I
`bx+ay=1+c`………ii
(ii) `+` (ii) `implies(a+b)x+(b+a)y=1+2c`
`implies (a+b)(x+y)=1+2c`
`impliesx+y=(1+2c)/(a+b)`………….iii
(i) `-` (ii) `implies(a-b)x+(b-a)y=-1`
`implies(a-b)(x-y)=-1`
`impliesx-y=-1/(a-b)`…………iv
(iii) `+` (iv) `implies2x=1/(a+b)+(2c)/(a+b)-1/(a-b)=1/(a+b)-1/(a-b)+(2c)/(a+b)`
`=(a-b-a-b)/((a+b)(a-b))+(2c)/(a+b)=(-2b)/(a^(2)-b^(2))+(2c)/(a+b)`
`:.x=c/(a+b)-b/(a^(2)-b^(2))`
(iii) `-` (iv) `implies2y=(1+2c)/(a+b)-1/(a-b)=1/(a+b)+(2c)/(a+b)+1/(a-b)`
`=1/(a+b)+1/(a-b)+(2c)/(a+b)=(a-b+a+b)/((a+b)(a-b))+(2c)/(a+b)`
`=(2a)/(a^(2)-b^(2))+(2c)/(a+b)`
`:.y=a/(a^(2)-b^(2))+c/(a+b)`
इस तरह `x=c/(a+b)-b/(a^(2)-b^(2))` और `y=c/(a+b)+a/(a^(2)-b^(2))`