Question

Solve for x, y, and z. Please show all steps.

Answer 1

Answer:

I have put my answer in the form (x,y,z)

One solution (3,2,4)

Another solution (-5,-4,-6)

Step-by-step 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)