+0  
 
0
491
3
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Solve the system of equations:

3a+b+c+d+e+f=43 
a+3b+c+d+e+f=51
a+b+3c+d+e+f=67
a+b+c+3d+e+f=32
a+b+c+d+3e+f=81
a+b+c+d+e+3f=52

 Feb 22, 2021
 #1
avatar+74 
+1

Using Wolfram Alpha, we get a = -104/253, b = 2080/253, c = 2496/253, d = 3328/253, e = 1508/253, f = 4056/253

 Feb 22, 2021
 #2
avatar+33614 
+3

You might have entered it into Wolfram Alpha incorrectly!  Check the last equation, for example:

Not 52.

 

Using Mathcad I get:

Alan  Feb 22, 2021
 #3
avatar+1224 
0

You can solve this without WolframAlpha. Start by adding all the equations together:

 

4a + 4b + 4c + 4d + 4e + 4f = 326

a + b + c + d + e + f = 81.5

 

If each equation is called (1), (2), (3), ..., etc, then you can subtract our "master equation" from each like so:

 

3a + b + c + d + e + f = 43

a + b + c + d + e + f = 81.5

 

Subtract, so 2a = -38.5 and a = -19.25.

 

The numbers may be incorrect, but you get the idea.

 Feb 22, 2021

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