Question 20906
x-y=3
x=y+3...or...y=x-3....
x^2 -10y-6=0 ,substituting y=x-3 from above eqn. in this we get ,
x^2-10(x-3)-6=0
x^2-10x+30-6=0
x^2-10x+24=0
find 2 factors for 24 which add up to -10.we find that -6*-4=+24 and -6-4=-10...so
x^2-6x-4x+24=0
x(x-6)-4(x-6)=0
(x-6)(x-4)=0
x-6=0 ....or...x-4=0
x=6...or...x=4