Question 1196448: I've been working on this for hours, but keep taking the wrong path. I need to solve the 4x4 equations:
2w + x - y + 2z = 10
-w - 3x + y - 5z = -11
3w + 5x - 3y + z = 35
-4w - 2x - 3y - z = 2
Found 4 solutions by ikleyn, MathLover1, MathTherapy, math_tutor2020: Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
2x1 + x2 - x3 + 2x4 = 10
-x1 - 3x2 + x3 - 5x4 = -11
3x1 + 5x2 - 3x3 + x4 = 35
-4x1 - 2x2 - 3x3 - x4 = 2
We rewrite the system of equations in matrix form and solve it using the Gauss method
2 1 -1 2 10
-1 -3 1 -5 -11
3 5 -3 1 35
-4 -2 -3 -1 2
Divide the 1st row by 2
1 0.5 -0.5 1 5
-1 -3 1 -5 -11
3 5 -3 1 35
-4 -2 -3 -1 2
to line 2 we add line 1 multiplied by 1; from line 3 we subtract line 1 multiplied by 3;
to line 4 add line 1 multiplied by 4
1 0.5 -0.5 1 5
0 -2.5 0.5 -4 -6
0 3.5 -1.5 -2 20
0 0 -5 3 22
Divide the 2nd line by -2.5
1 0.5 -0.5 1 5
0 1 -0.2 1.6 2.4
0 3.5 -1.5 -2 20
0 0 -5 3 22
from line 1 we subtract line 2, multiplied by 0.5; from line 3 subtract line 2 multiplied by 3.5
1 0 -0.4 0.2 3.8
0 1 -0.2 1.6 2.4
0 0 -0.8 -7.6 11.6
0 0 -5 3 22
Divide the 3rd line by -0.8
1 0 -0.4 0.2 3.8
0 1 -0.2 1.6 2.4
0 0 1 9.5 -14.5
0 0 -5 3 22
to line 1 we add line 3 multiplied by 0.4; to line 2 add line 3 multiplied by 0.2;
to line 4 add line 3 multiplied by 5
1 0 0 4 -2
0 1 0 3.5 -0.5
0 0 1 9.5 -14.5
0 0 0 50.5 -50.5
Divide the 4th line by 50.5
1 0 0 4 -2
0 1 0 3.5 -0.5
0 0 1 9.5 -14.5
0 0 0 1 -1
from line 1 we subtract line 4, multiplied by 4; from line 2 we subtract line 4, multiplied by 3.5;
from line 3 subtract line 4 multiplied by 9.5
1 0 0 0 2
0 1 0 0 3
0 0 1 0 -5
0 0 0 1 -1
x1 = 2
x2 = 3
x3 = -5
x4 = -1
Solved.
Answer by MathLover1(20850) (Show Source): Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website! I've been working on this for hours, but keep taking the wrong path. I need to solve the 4x4 equations:
2w + x - y + 2z = 10
-w - 3x + y - 5z = -11
3w + 5x - 3y + z = 35
-4w - 2x - 3y - z = 2
DO NOT bother to even look at that woman's RIDICULOUS method of solving this. As usual, it's complicated as HELL,
with all those solutions of variables that form UNNECESSARY fractions!! She needs to learn mathematics!
Looking at eqs (i) & (ii), it's obvious that "y" can be immediately eliminated by adding the two.
Likewise, looking at eqs (iii) & (iv), it's clear that "y" can also be immediately eliminated by subtracting one
equation from the other. So, let's do just that!
2w + x - y + 2z = 10 ----- eq (i)
- w - 3x + y - 5z = - 11 ----- eq (ii)
w - 2x - 3z = - 1 ---- Adding eqs (i) & (ii) ------ eq (v)
3w + 5x - 3y + z = 35 ----- eq (iii)
- 4w - 2x - 3y - z = 2 ----- eq (iv)
7w + 7x + 2z = 33 ----- Subtracting eq (iv) from eq (iii) ----- eq (vi)
- 3w - 9x + 3y - 15z = - 33 ----- eq (ii) ---- Multiplying eq (ii) by 3 ---- eq (vii)
3w + 5x - 3y + z = 35 ----- eq (iii)
- 4x - 14z = 2 ---- Adding eq (vii) & (iii)
------ Dividing by/Factoring out GCF, - 2
2x + 7z = - 1 ----- eq (viii)
w - 2x - 3z = - 1 ---- eq (v)
7w + 7x + 2z = 33 ----- eq (vi)
7w - 14x - 21z = - 7 ----- Multiplying eq (v) by 7 ---- eq (ix)
21x + 23z = 40 ---- Subtracting eq (ix) from eq (vi) ------ eq (x)
Now, we have 2 equations in 2 UNKNOWNS:
2x + 7z = - 1 ----- eq (viii)
21x + 23z = 40 ---- eq (x)
42x + 147z = - 21 ---- Multiplying eq (viii) by 21 ----- eq (xi)
42x + 46z = 80 ------ Multiplying eq (x) by 2 ----- eq (xii)
101z = - 101 --- Subtracting eq (xii) from eq (xi)
2x + 7(- 1) = - 1 ---- Substituting - 1 for z in eq (viii)
2x - 7 = - 1
2x = 6
w - 2(3) - 3(- 1) = - 1 ----- Substituting - 1 for z and 3 for x in eq (v)
w - 6 + 3 = - 1
w - 3 = - 1
- 2 - 3(3) + y - 5(- 1) = - 11 ----- Substituting - 1 for z, 3 for x, and 2 for w in eq (ii)
- 2 - 9 + y + 5 = - 11
- 6 + y = - 11

Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
The other tutors have great solutions, so I don't have much to add. Though I would like to mention relevant software tools.
For problems like this, the Linear Algebra Toolkit is an invaluable resource
The link is here
http://www.math.odu.edu/~bogacki/lat/
Click on "Enter". Then click on "Solving a system of linear equations"
We have m = 4 equations and n = 4 unknowns
The variables x1,x2,x3,x4 represent w,x,y,z in that order.
The boxes to the left of each represent the coefficients.
Type those coefficients in, as well as the right hand sides.
Once everything is typed in, double-check that the numbers have been entered correctly. Then hit the "submit" button.
This will provide the final answer. The step by step process is also provided. Click on "Show Details" to see the steps.
You should end up with
x1 = 2
x2 = 3
x3 = -5
x4 = -1
which means
w = 2
x = 3
y = -5
z = -1
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An alternative calculator is WolframAlpha
https://www.wolframalpha.com/
This is even easier to use since all you need to do is type the following shown in red
2w + x - y + 2z = 10,-w - 3x + y - 5z = -11,3w + 5x - 3y + z = 35,-4w - 2x - 3y - z = 2
Basically list each equation separated by commas.
WolframAlpha is great, but it doesn't show steps freely. You would need to become a paid subscriber to their service if you wanted to see the steps.
The first link (the linear algebra toolkit) is completely free for both steps and final answer.
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Another useful tool, that I personally use all the time, is GeoGebra.
https://www.geogebra.org/
It is completely free software. The drawback is that it doesn't show the step-by-step process.
To solve this system using GeoGebra, you would need to open the CAS (computer algebra system) panel.
Then in a CAS input box, you'll type the following shown in red
Solve[{2w + x - y + 2z = 10,-w - 3x + y - 5z = -11,3w + 5x - 3y + z = 35,-4w - 2x - 3y - z = 2}]
The input structure is very similar to WolframAlpha's input. Though we have "Solve" up front, and a pair of square brackets and curly braces surrounding the equations.
The template is Solve[{SYSTEM}] where "SYSTEM" is replaced with the system of equations in question.
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There are numerous other calculators and tools dealing with this topic. Feel free to explore to find your favorites.
Keep in mind that these tools are meant to help you, and not do your homework for you. You should still do the problems yourself to get practice.
When it comes to exams, it's likely your teacher will not allow such calculators to be used.
S/he would probably only allow things like a TI83 but may not even allow that.
In short, it's best to get practice doing the problem yourself and then using the calculators mentioned as a way to check your answer.
They are also helpful tools to see how to do something if you get stuck.
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