For the 2nd one, multiply top & bottom by xy to simplify it. Then what I'd do would be to substitute for either x or y from the first one, shake it down and then solve the resulting quadratic.
Been trying to jig this around for a while but i can't seem to get it. I know the basic method for simultaneous equation but this one with x and y as the denominator has thrown me off. Any help would be appreciated.
2/x + 3/y = 5
make x the subject in the seccond formula.
therefore ending up with:x=(4y)/(5y-3)
substitute for x in the first equation
after cancelling everything you end up with the quadratic 3y^2-11y+6=0
factorize it (3y-2)(y-3) =0
y= 3 and (2/3)
substitute in tiher equation to get corresponding x values
(x,y) = (0.5,3) , (4, 2/3)
but i am a little unsure about the fact that the highest power of both x and y in the two equations is 1 so by right to my knowledge they each should have just one solution, we usually eliminate the negative solution but since both solutions are positives in this one, i am unsure. But never the less the above answers work. (are correct)