Solve the system using the linear combination method.---(3x+2y -z =8), (-3x+4y+5z= -14),(x-3y+4z= -14).?

Question

How do you work the linear method out in this type of problem? I know how you work it out in a system of 2 problems, but I'm not sure about 3 equations.

Answer 1

LOOK:
1- 3X+2Y-Z=8
2- -3X+4Y+5Z=-14
3- X-3Y+4Z=-14
3- X= -14-4Z+3Y

1- 3*(-14-4Z+3Y)+2Y-Z=8
2- -3*(-14-4Z+3Y)+4Y+5Z=-14
got it?

Answer 2

What you need to do is eleminate the third variable first.
3x+2y-z=8
-3x+4y+5z=-14
add on both sides
3x-3x+2y+4y-z+5z=8-14
you get
6y+4z=-6

similarily add
-3x+4y+5z=-14
3*(x-3y+4z=-14)
you get

-5y+17z=-56

now you have two new equatuions
6y+4z=-6 and -5y+17z=-56
somlve these to get y and z. Put the values of y and z in any of the original equations to get x.

Answer 3

equation A: 3x+2y-z=8 equation B: -3x+4y+5z=-14 equation C: x-3y+4z=-14 upload A and B 6y + 4z = -6 (simplify) 3y + 2z = -3 multiply C by using three 3x -9y +12z = -40 two (now upload B) -3x +4y +5z = -14 ----------------------- -5y + 17z = -fifty six now you have 2 equations with 2 unknowns 3y + 2z = -3 (multiply by using 5) -5y + 17z = -fifty six (multiply by using three) 15y + 10z = -15 -15y + 51z = -168 ---------------------- 61z = -183 z = -3 3y +2(-3) = -3 3y -6 = -3 y = a million 3x +2(a million) -(-3) = 8 3x + 5 = 8 x = a million

Answer 4

3x+2y-z = 8 [EQ 1]
-3x+4y+5z = -14 EQ2]
x-3y+4z=-14 [EQ3]
Add EQ1 to EQ2 to get:
6y +4z = -6 [EQ4]
Multiply [EQ3] by 3 to get:
3x -9y +12z= -42 [EQ5]
Now add EQ5 to EQ2 to get:
-5y +17z = -56 [EQ6]
Multily EQ4 by 6 and EQ6 by 5 and add results together.
This gives 122z = -366, so z=-3.
Now put z= -3 into EQ6 and get:
-5y+(17)(-3) = -56, so -5y = =-5, so y=1
Now put z=-3 and y=1 into EQ1 to get:
3x +(2)(1) - (-3) = 8, so 3x+5 =8, 3x = 3, so x=1
The soluton is x=1, y = 1, z = -3.

Answer 5

It's the same way - you just have more variables to substitute for.

Answer 6

Work out your own homework.....And for the record, i have no idea