Question

Use the substitution method to solve the following system of equations.

3x + y – z = 2
2x + y + 2z = 8
2x – 3y + z = 21

(3, 4, 3)

(–3, 4, 3)

(3, –4, 3)

(3, 4, –3)

Answer 1

okay what is the asking about

Answer 2

Basically I need to figure out what x, y, and z is.

Answer 3

okay first take the easiest eqn with another and try to make cancellation till even get one parameter

Answer 4

Solve the first equation for z. Replace z in the second equation with that. Then you'll have an equation with just x and y. Solve that for Y.

Then replace z in the third equation with what you solved the first equation for. You'll have another equation with just x and y. Replace y with what you found above. Then you'll have an equation with just x. Solve for x and you'll have a number. Replace x in one of the x and y equations with that value to get y, then use one of the original equations to get z.

Show your work and we can help you figure it out.

Answer 5

K I"ll try right now.

Answer 6

So when I solve for z in the first equation I get, z=3x+y-2. Then I get 8x+3y=10 once I substiute the z thing in the 2nd equation.

Answer 7

Idk how I would get y alone in this equation right now. Would I do the same thing I did with the first and second, (isolate z). And do this with the 1st/3rd for y?

Answer 8

Oh wait i can cancel out this with the 3rd equation right?

Answer 9

You're off by a bit. Looks like you didn't multiply 2 by -2.

z=3x+y-2
2x + y + 2z = 8 -> Replace z:
2x + y + 2(3x+y-2) = 8
2x + y + 6x + 2y - 4 = 8
8x + 3y = 12

Answer 10

Oh okay yeah I just saw that. Thxx. Now how would I continue?

Answer 11

Solve that equation for y.

Answer 12

K. and that would be y=-8/3x+4

Answer 13

Doing the same thing, substituting z in the 3rd equation I got 5x-2y=23.