Question

questions in the pic

Answer 1

Answer:

x=1 and y=-1

Step-by-step explanation:

The given equations are:

$ax+by-a+b=0$                         (1)

$bx-ay-a-b=0$                          (2)

Now, multiplying equation (1) by b and equation (2) by a, we get

$abx+b^{2}y-ab+b^{2}=0$,

$abx-a^{2}y-ab-a^{2}=0$

Now, subtracting the above equations,

$(a^{2}+b^{2})y+(a^{2}+b^{2})=0$

$(a^{2}+b^{2})(y+1)=0$

$y=-1$

Putting the value of y=-1 in equation (1), we get

$ax+b(-1)-a+b=0$

$ax-b-a+b=0$

$a(x-1)=0$

$x=1$