Question 1205687
<pre>

Solve the system of linear equation below:
2w + 3x + y + 2z = 16
W + 4x + 3y + 3z = 15
3w + 2x + 4y + z = 22
W + 5x + 3y + 2z = 15

<font size = 4><font color = red><b>FYI:</font></font></b> {{{highlight_green(W <> w)}}}
This can be done by applying the system-of-equations solution, as follows:

2w + 3x +  y + 2z = 16 ------- eq (i)
 w + 4x + 3y + 3z = 15 ------- eq (ii)
3w + 2x + 4y +  z = 22 ------- eq (iii)
 w + 5x + 3y + 2z = 15 ------- eq (iv)

2w + 3x +  y + 2z = 16 ------- eq (i)
 w + 5x + 3y + 2z = 15 ------- eq (iv)
 w - 2x - 2y = 1 ------ Subtracting eq (iv) from eq (i) ----- eq (v)

3w + 2x + 4y +  z = 22 ------- eq (iii)
6w + 4x + 8y + 2z = 44 ------- Multiplying eq (iii) by 2 ---- eq (vi)
 w + 5x + 3y + 2z = 15 ------- eq (iv)
5w - x + 5y = 29 ------ Subtracting eq (iv) from eq (vi) ----- eq (vii)

3w + 2x +  4y +  z = 22 ------ eq (iii)
9w + 6x + 12y + 3z = 66 ------ Multiplying eq (iii) by 3 ---- eq (viii)
 w + 4x + 3y + 3z = 15 ------- eq (ii)
8w + 2x + 9y = 51 ----- Subtracting eq (ii) from eq (viii) ---- eq (ix)

 w - 2x - 2y =  1 ------ eq (v)
8w + 2x + 9y = 51 ------ eq (ix)
     9w + 7y = 52 ------ Adding eqs (v) and (ix) ----- eq (x)

 5w -  x +  5y = 29 ----- eq (vii)
10w - 2x + 10y = 58 ----- Multiplying eq (vii) by 2 ---- eq (xi)
 8w + 2x +  9y = 51 ----- eq (ix) 
     18w + 19y = 109 ---- Adding eqs (xi) and (ix) ----- eq (xii)

Finally, we have 2 equations in 2 variables:
      9w +  7y =  52 ----- eq (x)
     18w + 19y = 109 ----- eq (xii)
     18w + 14y = 104 ----- Multiplying eq (x) by 2 ---- eq (xiii)
            5y = 5 ---- Subtracting eq (xiii) from eq (xii)
           {{{highlight_green(matrix(1,5, highlight(y), "=", 5/5, "=", highlight(1)))}}}

        9w + 7y = 52 ---- eq (x)
      9w + 7(1) = 52 ---- Substituting 1 for y in eq (x)
             9w = 45
           {{{highlight_green(matrix(1,5, highlight(w), "=", 45/9, "=", highlight(5)))}}}

    5w - x + 5y = 29 ----- (vii)
5(5) - x + 5(1) = 29 ----- Substituting 1 for y and 5 for w, in eq (vii)
     25 - x + 5 = 29
       - x + 30 = 29
            - x = - 1
            {{{highlight_green(matrix(1,5, highlight(x), "=", (- 1)/(- 1), "=", highlight(1)))}}}

       3w + 2x + 4y + z = 22 ----- eq (iii)
 3(5) + 2(1) + 4(1) + z = 22 ----- Substituting 1 for y, 5 for w, and 1 for x in eq (1ii)
         15 + 2 + 4 + z = 22
                 21 + z = 22                   
                    {{{highlight_green(matrix(1,5, highlight(z), "=", 22 - 21, "=", highlight(1)))}}}

I never saw what Tutor @IKLEYN saw. Her solution is much, much shorter than mine because by using
subtraction and eqs (ii) and (iv), she was able to eliminate 2 of the 4 variables. Thus, use hers
if you're looking to solve this system by this method.</pre>