SOLUTION: solve by addition method. I am really good at adding 2 equations usually but 3 is hectic, everytime I do one the program tells me its wrong. 2x-y+z=1 3x-3y+4z=5 4x-2y+3z=4

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Question 1042629: solve by addition method. I am really good at adding 2 equations usually but 3 is hectic, everytime I do one the program tells me its wrong.
2x-y+z=1
3x-3y+4z=5
4x-2y+3z=4
Help would be very much appreciate. I am not getting it.
Found 3 solutions by ikleyn, josgarithmetic, MathTherapy:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
solve by addition method. I am really good at adding 2 equations usually but 3 is hectic, everytime I do one the program tells me its wrong.
2x-y+z=1
3x-3y+4z=5
4x-2y+3z=4
Help would be very much appreciate. I am not getting it.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

2x -  y + z  = 1   (1)
3x - 3y + 4z = 5   (2)
4x - 2y + 3z = 4   (3)

As a first step, multiply equation (1) by 2 and then distract from the equation (3).

Then complete it on your own.


Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
What addition method? Best method would be either Elimination or Matrix row reduction.

Start with this matrix:


First set of elementary row operations can be:
New R2 --- 2R2-3R1
New R3 --- R3-2R1
Just that much alone will put this matrix into upper triangular form, to this:


which shows a result for z already.

With that result, the next set of row operations should be:
New R1 ----- R1-R3
New R2 ----- R2-5*R3
which will give this matrix:



Obvious next step is multiply R2 by 1%2F21.

Obviously showing the result for value of variable y.


So far you have system%28z=2%2Cy=-1%2F7%29
and you still need to find x.

--
Take final shown row 1, substitute for y...
2x-1%2Ay=-1
2x-1%28-1%2F7%29=-1
x=-4%2F7-------which finishes the solution.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
solve by addition method. I am really good at adding 2 equations usually but 3 is hectic, everytime I do one the program tells me its wrong.
2x-y+z=1
3x-3y+4z=5
4x-2y+3z=4
Help would be very much appreciate. I am not getting it.
system%282x+-+y+%2B+z+=+1%2C+3x+-+3y+%2B+4z+=+5%2C+4x+-+2y+%2B+3z+=+4%29

Multiply eq (i) by - 2. Add this new equation to eq (iii) to ELIMINATE x and y, and get a value for z

Substitute the value for z into eq (ii) and simplify to get an equation in variables x and y

Substitute the value for z into eq (i), and subtract this equation in variables x and y from the above equation, in x and y to get the value of x

Substitute values for x and z into any of the 3 ORIGINAL equations to get the value of y.

Voila!! All done!



By the way, you should get: highlight_green%28system%28x+=+0%2C+y+=+1%2C+z+=+2%29%29