Question 1205687
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Solve the system of linear equation below:

<pre>
2w + 3x + y + 2z = 16    (1)
W + 4x + 3y + 3z = 15    (2)
3w + 2x + 4y + z = 22    (3)
W + 5x + 3y + 2z = 15    (4)
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<pre>
From equation (2), subtract equation (4).  You will get

    -x      + z = 0,  or  x = z.    (5)


Based on (5), replace z by x everywhere in equations from (1) to (3)  
and combine like terms. You will get then

    2w + 5x +  y = 16     (6)
     w + 7x + 3y = 15     (7)
    3w + 3x + 4y = 22     (8)


Add equations (6) and (7) to get new equation (9).  Keep equation (8) as is. You will get then

    3w + 12x + 4y = 31    (9)
    3w +  3x + 4y = 22    (8)


Now subtract equation (8) from equation (9).  You will get

          9x     =   9;   hence,  x = 9/9 = 1.


Then from (5) we have  z= 1,  too.  So, the solutions for x and z are just found.


Next, substitute  x= 1 into equations (6) and (7).  You will get then

    2w + 5 +  y = 16    (6')
     w + 7 + 3y = 15    (7')


or, collecting constant terms in the right side

    2w +  y = 11         (6")
     w + 3y =  8         (7")


Now multiply equation (7") by 2 and subtract equation (6") from it.  You will get

         5y = 5,  or  y= 1.


As a final step, substitute y= 1 into equation (6") and get

      2w + 1 = 11,  or  2w = 11-1 = 10,  w = 10/2 = 5.


<U>ANSWER</U>.  w= 5,  x= 1,  y= 1,  z= 1.
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Solved.


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When you use outside solver, it is a good way to get the answer quickly or to check your solution, 
but you will learn nothing from it, except of useful information on existing a relevant solver in the Internet.


Hope my solution will inspire you for searching your own way to solve the problem.


As soon as you notice that equations (2) and (4) are very similar and allow elimination of several terms, 
the next path is to reduce the size of the system to 3 equations, 2 equations and 1 equation, step by step.