SOLUTION: 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

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Question 1205687: 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

Found 4 solutions by Theo, MathLover1, MathTherapy, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
this is a lengthy process, best solved by use of a gaussian elimination calculator, such as the one found at https://onlinemschool.com/math/assistance/equation/gaus/

the process is to have a 1 on each row for different variables on each row.

here's a picture of the process as it goes through row manipulation to achieve that effect.

the whole idea is to make the matrix have a 1 in the position for each variable on individual rows.

the 4th display from the bottom shows you what i mean.

this is done through row manipulation.

here's a referemce.

https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom)/02%3A_Matrices/2.02%3A_Systems_of_Linear_Equations_and_the_Gauss-Jordan_Method

here's a youtube by patrickJMT that might help.

https://www.youtube.com/watch?v=Ey62H_oaqoE

patrick has several videos on the subject, as do other tutors on youtube.

the process is logical, but painful to execute.

i've tried it manually several times.
it's prone to errors and can lead you down dead ends, requiring you to start again.

try the calculator yourself to see how it works for you.

if you want to tackle it manually, pick a simple problem to start until you get the feel for it.

something with two rows and 2 variagles to solve, then maybe 3 rows with 3 variable to solve.

this one is more complex with 4 variables and 4 rows to solve.

Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!

using Gauss-Jordan elimination calculator
https://matrix.reshish.com/gaussSolution.php

Your matrix

Find the pivot in the 1st column and swap the 2nd and the 1st rows

Multiply the 1st row by

Subtract the 1st row from the 2nd row and restore it

Multiply the 1st row by

Subtract the 1st row from the 3rd row and restore it

Subtract the 1st row from the 4th row

Find the pivot in the 2nd column and swap the 4th and the 2nd rows

Multiply the 2nd row by

Subtract the 2nd row from the 1st row and restore it

Multiply the 2nd row by

Subtract the 2nd row from the 3rd row and restore it

Multiply the 2nd row by

Subtract the 2nd row from the 4th row and restore it

Make the pivot in the 3rd column by dividing the 3rd row by

Multiply the 3rd row by

Subtract the 3rd row from the 1st row and restore it

Multiply the 3rd row by

Subtract the 3rd row from the 4th row and restore it

Make the pivot in the 4th column by dividing the 4th row by

Multiply the 4th row by

Subtract the 4th row from the 1st row and restore it

Multiply the 4th row by

Subtract the 4th row from the 2nd row and restore it

Multiply the 4th row by 18/5

Subtract the 4th row from the 3rd row and restore it

Solution set:

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!

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.

Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website!
.
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)
~~~~~~~~~~~~~~~~~~

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.