(1)/(a+b+x)=(1)/(a)+(1)/(b)+(1)/(x),x!=0,-(a+b)

Long-answer type questions (L.A.) (1)/(a+b+x)=(1)/(a)+(1)/(b)+(1)/(x){xne0,-(a+b)}

Solve (1)/(a+b+x)=(1)/(a)+(1)/(b)+(1)/(x),a+b!=0

Solve the following quadratic equation by factorization method (1)/(a+b+x)=(1)/(a)+(1)/(b)+(1)/(x),a+b!=0

Solve the following quadratic equations by factorization methd: (1)/(a+b+x)=(1)/(a)+(1)/(b)+(1)/(x),a+b!=0

Solve for x : (1)/(a + b + x) = (1)/(a) + (1)/(b) + (1)/(x) , a ne b ne 0 , x ne 0 , x ne -(a + b)

Solve for 'x' : (1)/(a+b+x)=(1)/(a)+(1)/(b)+(1)/(x) " " a != 0, b!=0, x !=0

Solve: (1)/((a+b+x))=(1)/(a)+(1)/(b)+(1)/(x),[xne0,xne-(a+b)].

(1)/(x+a)+(1)/(x+b)=(1)/(x+a+b)+(1)/(x)

Long-answer type questions (L.A.) (1)/(x)-(1)/(x+b)=(1)/(a)-(1)/(a+b)(xne0,-b)

If the lines a x+y+1=0,x+b y+1=0, and x+y+c=0(a , b , c being distinct and different from 1) are concurrent, then (1/(1-a))+(1/(1-b))+(1/(1-c))= (a)0 (b) 1 (c) 1/((a+b+c)) (d) none of these