easy, 3 and 0.5
but ok the math way
xy=1.5
x+y=3.5
y=3.5-x
x(3.5-x)=1.5
3.5x-x^2=1.5
0=x^2-3.5x+1.5
quadratic
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} [/tex]
a=1
b=-3.5
c=1.5
x=[tex] \frac{-(-3.5)+/- \sqrt{(3.5)^{2}-4(1)(1.5)} }{2(1)} [/tex]
x=[tex] \frac{3.5+/- \sqrt{12.25-6} }{2} [/tex]
x=[tex] \frac{3.5+/- \sqrt{6.25} }{2} [/tex]
x=[tex] \frac{3.5+/- 2.5}{2} [/tex]
x=[tex] \frac{3.5+ 2.5}{2} [/tex] or x=[tex] \frac{3.5- 2.5}{2} [/tex]
x=6/2 or 1/2
x=3 or 0.5
those are the numbers
3 and 0.5
