Question 62306: I am having trouble solving these systems. I get lost somewhere in the middle and it becomes a mess. Please help.
x-2y-3z=4
2x-4y+5z=-3
5x-6y+4z=-7
and
2x-y=-1
-2x+z=1
y-z=0
Thank you.
Found 2 solutions by ankor@dixie-net.com, ptaylor:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! You can solve this as a matrix, but you can also elimination and substitution:
:
x - 2y - 3z = 4
2x -4y + 5z = -3
5x -6y + 4z = -7
:
Let's just deal with the 1st & 2nd equation:
x - 2y - 3z = 4
2x -4y + 5z = -3
:
Mult the 1st equation by 2, leave the 2nd equation as it is:
2x - 4y - 6z = 8
2x - 4y + 5z = -3
---------------------subtract & we eliminate x & y, it's easy to find z!
0x + 0y -11z = 11
z = 11/-11
z = -1
:
Now that we know z = -1 we can rewrite the 3 equations:
x - 2y - 3(-1) = 4 >> x -2y + 3 = 4 >> x - 2y = 4 - 3
2x -4y + 5(-1) = -3 >> 2x -4y - 5 = -3 >> 2x -4y = -3 + 5
5x -6y + 4(-1) = -7 >> 5x -6y - 4 = -7 >> 5x - 6y = -7 + 4
We end up with 3 equation with 2 unknowns:
x - 2y = +1
2x -4y = +2
5x -6y = -3
:
Use the 1st equation for substitution:
x = 2y + 1
:
Substitute for x in the 3rd "2 unknown" equation:
5(2y+1) - 6y = -3
10y + 5 - 6y = -3
4y = -3 - 5
4y = -8
y = -8/4
y = -2
:
Find x using x = 2y +1
x = 2(-2) + 1
x = -4 + 1
x = -3
:
We have x = -3, y = -2; z = -1
Choose one of the original equations to check our solutions, the 2nd equation:
2x -4y + 5z = -3
2(-3) - 4(-2) + 5(-1) = -3
-6 + 8 - 5 = -3; equality reigns
:
:
:
and
2x - y = -1
-2x + z = 1
y - z = 0
:
Look at the equation y - z = 0, that means y = z, right,
Let's replace y with z in the 1st equation and pair it with the 2nd equation:
2x - z = -1
-2x + z = 1
------------ add these and you get
0 + 0 = 0
:
That means there is no unique solution to this system
Answer by ptaylor(2198)  (Show Source):
You can put this solution on YOUR website! To solve:
(1) x-2y-3z=4
(2) 2x-4y+5z=-3
(3) 5x-6y+4z=-7
First, I will state the operations that we are allowed to perform on these equations without changing the solutions:
(a) We can interchange any two equations
(b) We can multiply an equation by a constant other than zero
(c) We can multiply an equation by a constant other than zero and add it to another equation, replacing that equation
We'll start by doing (c). We'll multiply equation (1) by -2 and then add equation (1) to equation (2) and we will get:
(1) -2x+4y+6z=-8
(2) 0x+0y+11z=-11 from this equation, we see that z=-1
(3) 5x-6y+4z=-7
Next, we'll plug z=-1 into eq (1) and (3) and get:
(1) -2x+4y=-2
(3) 5x-6y=-3
Next, we'll multiply eq (1) by 5 and eq (3) by 2 and we get:
(1) -10x+20y=-10
(3) 10x-12y=-6
Now add eq (1) to eq (3), replacing eq (3) and we have:
(1) -10x+20y=-10
(3) 0x+ 8y=-16 and here we have y=-2
Next divide eq (1) by -10
(1) x-2y=1 now plug in y=-2 and we have:
(1) x+4=1 ;x=-3
We now have
x=-3
y=-2
z=-1
I'll leave the second problem up to you. It looks to be fairly straightforward. Hint: You know from eq (3) that y=z. What happens if you plug that into equation (1), then add eq (1) and (2)?
Hope this helps----ptaylor
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