x-2y+3z=7,2x+y+z=4,-3x+2y-2z=-10 solve by subsiution

Question

I would like to learn to solve the problem by substitution rather than elimination

Answer 1

x-2y+3z=7         equation 1

2x+y+z=4          equation 2

-3x+2y-2z=-10   equation 3

 

Use one of the equations to solve for 1

variable in terms of the others.  From

equation 2 we have:

 

y=-2x-z+4

 

Substitute that into the other two equations

 

 x-2(-2x-z+4)+3z=7.   equation 1

x+4x+2z-8+3z=7

5x+5z-8=7

5x+5z=15

x+z=3

 

-3x+2(-2x-z+4)-2z=-10.   equation 3

-3x-4x-2z+8-2z=-10

-7x-4z+8=-10

-7x-4z=-18

 

We now have 2 equations with 2 unknowns:

 

x+z=3.          equation 1a

-7x-4z=-18.   equation 3a

 

From x+z=3 we have:

z=-x+3

Plug that into equation 3a

 

-7x-4(-x+3)=-18

-7x+4x-12=-18

-3x=-6

x=2

 

since x+z=3 and x=2 we have z=1

 

We now have values of x and z.

Plug those into one of the original equations.

 

Using equation 1:  x-2y+3z=7

2-2y+3(1)=7

-2y+5=7

-2y=2

y=-1

 

x=2, y=-1, z=1

 

If you substitute those values into the

original 3 equations you will see that they

work for the stated equations.

 

Hope this helps.  Good luck.

Answer 2

From the first equation:

x=7+2y-3z

plug into the second:

2(7+2y-3z) +y+z=4

14+4y-6z+y+z=4

5y-5z=-10

y-z=-2

y=z-2

plug into x=7+2y-3z:

x=7+2z-4-3z = 3-z

So plug the equations of x and y as a function of z into the third one:

-3(3-z)+2(z-2)-2z=-10

-9+3z+2z-4-2z=-10

3z=3

z=1

so x=3-z=3-1=2

y=z-2=1-2=-1

So x=2,y=-1,z=1