Solve this system of equations

Question

5x+3y+z=0

 

x-3y+2z=9

 

14x-2y+3z=1

Answer 1

By rearranging 5x+3y+z=0, we get: 3y=-5x-z. 

By rearranging x-3y+2x=9, we get: 3y=-9+x+2z.

Now, we can set these equations equal to each other since the left sides of the equations are the same. 

So we get: -5x-z=-9+x+2z.

And we can solve for x in terms of z: x=(9-3z)/6=(3-z)/2.

 

Now, we want to get y in terms of z. Let's arbitrarily use the first equation, now substituting x for (3-z)/2. 

5(3-z)/2 +3y+z=0.

-3y=(15/2)-(5/2)z + z = (15/2) +(3/2)z

So y = ((15/2)+(3/2)z)/-3 = (-5/2)-(1/2)z=(-5-z)/2. 

 

Now that y and x are in terms of z, we can plug both of these values into the last equation to get z:

 

1=14x-2y+3z=14((3-z)/2)-2((-5-z)/2)+3z

=7(3-z)+5+z+3z

=21-7z+5+z+3z

=26-3z

 

-25=-3z, so z=25/3.  

 

Now, we can plug 25/3 back into the equations for x and y in terms of z to find the values of x and y. 

x=(3-z)/2=(3-25/3)/2=(-16/(3*2)=-8/3.

y=(-5-z)/2=(-5-25/3)/2=-40/(3*2)=-20/3.