Question

Alright, so thanks to a couple of wonderful people - I DO have somewhat of a better understanding of systems of equations with 3 variables. BUT these ones threw me off. Can someone please help? (Elimination/Substitution)
x-z=5
y+3z=12
2x=y=7

AND

x+y=5
3x+z=2
4y-z=8

Answer 1

What's the third equation of the first system?

Answer 2

Let's start with the second system and use substitution.
Solve the first equation for y:
x + y = 5
y = 5 - x
Now solve the second equation for z:
3x + z = 2
z = 2 - 3x
Now that you know what y and z are interms of x, plug those results into the third equation:
4y - z = 8
4(5 - x) - (2 - 3x) = 8
Solve for x:
20 - 4x - 2 + 3x = 8
18 -x = 8
-x = -10
x = 10
Now substitute x into the original first equation:
x + y = 5
10 + y = 5
y = -5
Now substitue x into the original second equation:
3x + z = 2
3(10) + z = 2
30 + z = 2
z = -28
The solution is: x = 10, y = -5, z = -28

Answer 3

oh sorry! The first set, 3rd equation is 2x+y=7

Answer 4

Since the "+" character and the "=" character are on the same key on the keyboard, I'm going to assume that the third equation of the first system is: 2x + y = 7.
We have this system then:
x - z = 5
y + 3z = 12
2x + y = 7

Answer 5

Yes, I figured it out.

Answer 6

Thank you so much! You have no idea how helpful you are!

Answer 7

Let's use elimination on the first system.
Multiply the first equation by -2 and add to the third equation:
-2x + 2z = -10
2x + y = 7
-------------------
y + 2z = -3
Notice now that the original second equation also has only the two variables y and z.
Rewrite this new equation and below it, multiply the original second equation by -1. Then add them:
y + 2z = -3
-y - 3z = -12
______________
-z = -15
z = 15
Now substitute value of z into first original equation:
x - z = 5
x - 15 = 5
x = 20
Now substitute z = 15 into second original equation:
y + 3z = 12
y + 3(15) = 12
y + 45 = 12
y = -33
Solution is: x = 20, y = -33, z = 15

Answer 8

You are wonderful! Seriously thank you!