Find the values of `a` and `b` for which the simultaneous equations `x+2y=1` and `(a-b)x+(a+b)y=a+b-2` have infinitely many solutions.

Explanation

The condition for simultaneous equations having infinitely many solution is `(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))` ……………..1
For `x+2y=1,a_(1)=1, b_(1)=2, c_(1)=1`
For `(a-b)x+(a+b)y=a+b=2`
`a_(2)=a-b,b_(2)a+b,c_(2)=a+b-2`
Substituting these values in (1)
`1/(a-b)=2/(a+b)=1/(a+b-2)`
Now `1/(a-b)=1/(a+b-2)`
`:.a+b-2=a-b :. b-2=-b :. b+b=2`
`:.2b-2 :. b=2/2 :.b=1`
Substituting `b=1`
`1/(a-b)=2/(a+b)`
`:.1/(a-1)=2/(a+1) :. a+1=2(a-1)`
`:. a+1=2a-2 :. a-2a=-2-1`
`-a=-3 :.a=(-3)/(-1):.a=3`
Ans The values of a and b are 3 and 1 respectively.