xy=30
(10.3-y)y=30
y^2-10.3y+30=0
Use the quadratic equation.
Notice the serd: b^2-4ac is negative, so there are no real solutions.
I wonder if you meant x/y=30
orx=30y
30y+y=10.3
y=10.3/31
if x.y=30
and x+y=10.3
what is x ?
1 answer