[tex] \frac{3}{2} = \frac{2x}{2x+y} [/tex]
From any proportion, we get another proportion by inverting the extremes (or the means):
[tex] \frac{2x+y}{2} = \frac{2x}{3} [/tex] = k
so we have:
2x=3k
2x+y=2k therefore:
3k+y=2k
y= - k
x=[tex] \frac{3}{2} k[/tex]
[tex] \frac{x}{y} [/tex] = - [tex] \frac{3}{2} [/tex]
The corect answer is A. -3/2
or:
[tex] \frac{3}{2} = \frac{2x}{2x+y} [/tex]
From any proportion, we get another proportion by inverting the extremes and the means:
[tex] \frac{2x+y}{2x} = \frac{2}{3} [/tex]
We use a property of proportions:
[tex] \frac{a}{b} = \frac{c}{d} [/tex] where a, d are extremes and b,c are means and the product of the extremes equals the product of the means (a*d=b*c),
so we have
[tex] \frac{a-b}{b} = \frac{c-d}{d} [/tex] or
[tex] \frac{a+b}{b} = \frac{c+d}{d} [/tex] (you can check this also by "the product of the extremes equals the product of the means")
[tex] \frac{2x+y}{2x} = \frac{2}{3} [/tex]
[tex] \frac{(2x+y)-2x}{2x} = \frac{2-3}{3} [/tex]
[tex] \frac{y}{2x} = \frac{-1}{3} [/tex]
3y = - 2x
[tex] \frac{x}{y} = -\frac{3}{2} [/tex]
