Maths
posted by Smangele Phakathi .
Can you pliz help me solve the following simultaneous equation 3xy=2 and 3y+9x=4

3xy=2 Multiplt it by 3
9x3y=6 Add to it the other eq.
9x+3y=4 You get 18x=10 or x=5/9
Substitute in any eq.
3*5/9y=2 or y=15/92=5/3 2
or y=(56)/2=1/6.
Check: 3*5/9(1/6)=2 or 5/3+1/6=2
or (10+2)/6=2 ok.
9*5/93*1/3=4 or 51=4 ok. 
or ... using substitution
from the 1st: y = 3x2
sub into the 2nd
3(3x2) + 9x = 4
9x  6 + 9x = 4
18x = 10
x = 5/9
back into the 1st:
y = 3(5/9)  2
= 5/3  2
= 1/3
in original:
1st equation:
LS = 3(5/9)  (1/3) = 6/3 = 2 = RS
2nd equation:
LS= 3(1/3) + 9(5/9) = 1 + 5 = 4 = RS
So x = 5/9 and y = 1/3