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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. |
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Answer» 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. |
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