I need help with these simultaneous linear equations?

Question

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


2x-5y=1
3x+2y=11

Answer 1

x+y+2z=-7.....1)
x+ 4z=-7......2)
-y -z=3...........3)
substitute from 3): y = 3 -z in 1):

x + 3 -z +2z = -7
=> x +z = -7-3=-10.....4)
multiply both sides by 4:
=> 4x + 4z = -40 .....5)
x +4z = -7 .........2)
subtract 2)from 5):

3x = -40 +7 = -33
=> x= -11
substitute in 2)
-11 +4z =-7
=> 4z = -7 +11 = 4
=> z =1
substitute in 3)
-y-(-1) =3
-y +1 = 3
-y = 3 -1 =2
=> y= -2

similarly solve the second question

Answer 2

Another answerer has given you the steps to solve these LE's. But you should know why those are the steps to take.

In general, we first solve SLE's by finding what one variable equals in terms of another variable in the equations. For example, 2x = 1 + 5y from your second set of SLE's. In which case x = (1 + 5y)/2 is x in terms of y.

Second we plug x in terms of y into one of the other remaining eqns in the SLE's. In this example, 3x + 2y = 11, is the other eqn in the SLE's so we put (1 + 5y)/2 into wherever there is an x.

Therefore, 3(1 + 5y)/2 + 2y = 11; so that 3(1 + 5y) + 4y = 22 = 3 + 15y + 4y = 3 + 19y and 19y = 19, so y = 1.

When y = 1, 2x - 5y = 1 becomes 2x - 5*1 = 1 and 2x = 6, so that x = 3. So now we have both x and y; their values can be plugged into either of the two eqns in the SLE.

CHECK YOUR WORK by plugging both x and y values you found into any one of the two eqn in the SLE's. For example, 3x + 2y = 11 = 3*3 + 2*1 = 11, so y = 1 and x = 3 are valid answers.

Answer 3

let me answer the second one first
multiply the equation {2x-5y=1}with 3
=> 6x-15y=3 ---(1)
multiply the equation{3x+2y=11}with2
=>6x+4y=22 ---(2)
now subtract (2) from (1)
=>{-19y=-19}
=>y=1
substitute this in any equation
=>x=3


now the first one
from the third eq,
y=(-3-z)
put this in first eq,
=>x+z=-4
and subtract it from x+4z=-7
thus u get
x=(-3)
z=(-1)
y=(-2)

Answer 4

1)
From eq2: x= -7-4z
From eq3: y= -z-3
Plug results into eq1: (-7-4z)+(-z-3)+2z=7. So, z= -17/3.
No plug the value of z into eq 2, to solve for 'x' and then eq3 to solve for y...

2)same thing..
From eq1 write x in terms of y. Plug into eqn 2 and solve. Once you find 'y' from eqn2, plug into eq1 to find 'x'.

Answer 5

The correct answers are x = -3, y = -2, z = -1, with Matlab.