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
FYI:
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)
9w + 7y = 52 ---- eq (x)
9w + 7(1) = 52 ---- Substituting 1 for y in eq (x)
9w = 45
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
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
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.
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Solve the system of linear equation below:
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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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.
ANSWER. w= 5, x= 1, y= 1, z= 1.
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.
