1/[x+3] + 6 / [x^2-9]=1
but x^2-9=[x+3][x-3]
1/[x+3] +6/{ [x+3][x-3]} = 1
multiply both sides of the = sign by [x+3][x-3]
[x+3][x-3] / [x+3] + 6[x+3][x-3]/[x+3][x-3] simplify. [ x not eual to -3 or +3]
[x-3]+6=[x+3][x-3] subtract x-3 from both sides of = sign
6=[x^2-9]-[x-3] clear brackets
6=x^2-x-6 subtract 6 from each side of = sign
x^2-x-12=0 factor
factors of 1= 1.1
factors 12=1,12 2,6 3,4
we want a set whose difference of cross products = 1
1,1 and 3,4 does it
0=[x-4][x+3]
x=4 answer x=-3 is ambiguous answer
Arthur
