If this is a system you have Y expressed in terms of X.
2(3X - 5) - X = 0
6X - 10 - X = 0
5X = 10
X = 2
--------------find Y
Y = 3(2) - 5
Y = 6 - 5
Y = 1
----------------check in first equation
2(1) - 2 = 0
2 - 2 = 0
0 = 0
both check and system is consistent
2y + 2xy' + 4y + 4xy' = 0 6y + 6xy' = 0 y' = -y/x
For y - 2y - 3y equals 0, y equals 0.
a = 3 and y = 0
Given that the first y value is 0 (and not 400), it is very difficult. Here is one possibility: y = -5/9x6 + 11 2/3x5 - 97 2/9x4 + 408 1/3x3 - 902 2/9x2 + 1080x
x = y = 0 ?
The y-intercept equals 0.
y = 0x = y + 2plug 0 in for yx = 0 + 2x = 2
0
A pair of simultaneous eq;ns. 3x + y = 4 x + y = 0 Subtract the two eq'ns. This will eliminate 'y' 2x = 4 x = 2 Substitute this value of 'x' into either eq'n for 'y' 2 + y = 0 y = -2 or 3)2) + y = 4 6 + y = 4 y = 4 - 6 y = -2 ( again) . So the answer is ( x,y) = ( 2, -2).
If ( x = 0 ) and ( y = 1 ), then ( xy = 0 \times 1 = 0 ). Therefore, the value of ( xy ) is 0.
y = 0" the answer is YES.
maybe