Answer:
I have put my answer in the form (x,y,z)
One solution (3,2,4)
Another solution (-5,-4,-6)
Explanation:
I'm going to try to do this by a bunch of substitution.
I'm going to solve first equation for x, second for y, and third for z.
Commutative property x+xy+y=11
Distributive property x(1+y)+y=11
Subtraction property x(1+y)=11-y
Division property x=(11-y)/(1+y)
I'm going to do the other 2 equations in a similar way:
So the second equation solving for y: y=(14-z)/(1+z)
The third equation solving for z: z=(19-x)/(1+x)
I'm going to plug my new first equation into my third equation giving me:
z=(19-[(11-y)/(1+y)])/(1+[(11-y)/(1+y)]
Now I'm going to clean this up by multiplying by compound fraction by (1+y)/(1+y).
z=(19(1+y)-(11-y)]/[1(1+y)+(11-y)]
z=[19+19y-11+y]/[1+y+11-y]
z=[8+20y]/[12]
Simplify
z=(2+5y)/3
Now I'm going to sub this into my non-rewrite of equation 2:
y+(2+5y)/3+y(2+5y)/3=14
Multiply both sides by 3 to clear fractions
3y+(2+5y)+y(2+5y)=42
3y+2+5y+2y+5y^2=42
5y^2+10y+2=42
Subtract 42 on both sides
5y^2+10y-40=0
Divide both sides by 5
y^2+2y-8=0
Factor
(y+4)(y-2)=0
So y=-4 or y=2
If y=-4 then x=(11-(-4))/(1+(-4))=15/-3=-5 and z=(2+5*-4)/3=-18/3=-6
So one solution is (-5,-4,-6)
If y=2 then x=(11-2)/(1+2)=9/3=3 and z=(2+5*2)/3=12/3=4
So another solution is (3,2,4)