2x + 5y = 2-3x - y = -3from the second equation:-y = -3 + 3xy = 3 - 3xsubstitute this value in the first equation:2x + 5(3-3x) = 22x + 15 - 15x = 2-13x + 15 = 2-13x = 2 - 15-13x = -13so, x = 1Now, solve for y :y = 3 - 3xy = 3 - 3 (1)y = zero
y-3x+6y=12 7y=3x+12 y=3x/7 +12/7
3x + 2y = 12 ie 2y = 12 - 3x so y = 6 - 3x/2
3xy + 3y2 = 3y (x + y)
x = 12
-2x-10y-12+3xy=4Put all x's to the other side and the ones with y's on the other:-10y+3xy=4+2x+12-10y+3xy=16+2xNow you factor a y from the side with the y's:y(-10+3x)=16+2xDivide both sides by (-10+3x):y=(16+2x)/(-10+3x)
yes 3xy=6
2x + 5y = 2-3x - y = -3from the second equation:-y = -3 + 3xy = 3 - 3xsubstitute this value in the first equation:2x + 5(3-3x) = 22x + 15 - 15x = 2-13x + 15 = 2-13x = 2 - 15-13x = -13so, x = 1Now, solve for y :y = 3 - 3xy = 3 - 3 (1)y = zero
M=3 because y+y+y= 3xY and MY= MxY
y-3x+6y=12 7y=3x+12 y=3x/7 +12/7
You plug in the values of x and y and solve 12x + 16y = ? 12(1) + 16(2) = 12 + 32 = 44
3x + 2y = 12 ie 2y = 12 - 3x so y = 6 - 3x/2
-y + y equals 0.
1,3,x,y,3x,3y,xy,3xy
3xy
in order to solve this problem you have to solve for x first wich you are given the data to pre forme so x=4 and x.2=8 and 8=y and y-4=4=a=x so a+8=12 and 12=b
3xy + 3y2 = 3y (x + y)