I need some help!!!!!! solve by substitution: x - y + 2z = -7, y + z = 1, and x - 2y - 3z = 0.

Question

x - y + 2z = -7

y + z = 1

x - 2y - 3z = 0

 

Write your answer as a point, an ordered triple: (x,y,z), without spaces

Answer 1

Subtract the third statement from the first.

 

You should get:

x-y+2z-(x-2y-3z)=-7-0

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

 

y+5z=-7

 

 

If you subtract that above statement from the second statement (y+z=1), you should get:

 

y+z-(y+5z)=1- -7

y+z-y-5z=8

-4z=8

z=8/-4

z=-2

 

So, z=-2

 

If you substitute that value of y into the second statement (y+z=1),  you get:

 

y-2=1

y=1+2

y=3

 

So, y=3

 

Finally if you substitute those values into x-y+2z=-7, you get

x-3+2(-2)=-7

x-3-4=-7

x-7=-7

x=-7+7

x=0

 

 

x=0

 

So the final answer is 

 

(0,3,-2)

 

Let me know if you have more questions

Hope this helps!

Answer 2

(1)     x   - y + 2z = -7

(2)            y  + z  = 1

(3)     x - 2y  - 3z = 0

 

Solve (2) for z.

 

z = 1 - y

 

Plug it into (1) and (3)

 

x - y + 2(1 - y) = -7   ⇒   x - y + 2 - 2y = -7   ⇒   x - 3y = -9

 

x - 2y - 3(1 - y) = 0   ⇒   x - 2y - 3 + 3y = 0   ⇒   x + y = 3

 

So now we have the system

 

(4)       x - 3y = -9

 

(5)       x + y = 3

 

Solve (5) for y

 

y = 3 - x

 

Plug this into (4)

 

x - 3(3 - x) = -9

 

x - 9 + 3x = -9

 

4x = 0

 

x = 0

 

Now back-substitute

 

y = 3 - x = 3 - 0 = 3

 

z = 1 - y = 1 - 3 = -2

 

So the solution is (x,y,z) = (0,3,-2)

Answer 3

x-y+2z=-7

y+z=1

x-2y-3z=0

y+z=1 implies that z=1-y

substitute

x-y+2(1-y)=-7

x-2y-3(1-y)=0

 

x-y+2-2y=-7

x-2y-3+3y=0

 

x-3y+2=-7

x+y-3=0

 

x-3y=-9

x+y=3

 

subtract

-4y=-12

y=3

3+z=1

z=-2

 

x-2(3)-3(-2)=0

x-6+6=0

x+0=0

x=0,or...

 

x-3+2(-2)=-7

x-3-4=-7

x-7=-7

x=0 again

x=0, y=3, z=-2

(0,3,-2)

 

 

 

Answer 4

From the second equation:

y=1-z

plug that into the first equation:

x-(1-z)+2z=-7

x-1+z+2z=-7

x=-6-3z

plug both y-1-z and x=-6-3z into the third equation:

-6-3z-2(1-z)-3z=0

-6-3z-2+2z-3z=0

-4z=8

z=-2

x=-6-3*(-2)=0, y=1-z=1+2=3

So (0,3,-2)