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Suppose a, b, c, and d are real numbers which satisfy the system of equations

a + 2b + 3c + 4d = 10

4a + b + 2c + 3d = 4

3a + 4b + c + 2d = 10

2a + 3b + 4c + d = −4.

Find a + b + c + d.

 Oct 25, 2020
 #1
avatar
+1

We write the equation in matrix form

 

[1 2 3 4][a][10]

[4 1 2 3 ][b][4]

[3 4 1 2][c][10]

[2 3 4 1][d][-4]

 

Solving, we get a = -21/20, b = 39/20, c = -71/20 and d = 89/20, so a + b + c + d = 9/5

 Oct 25, 2020
 #2
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+3

Add the four equations together and we get

10a + 10b + 10c + 10d = 20,

so

a + b + c + d = 2.

 Oct 26, 2020
 #5
avatar+111600 
+2

I like this answer guest.    laugh

 

Heureka has elaborated on it below....   for anyone who doesn't already understand.   laugh

Melody  Oct 26, 2020
 #3
avatar+25597 
+5

Suppose a, b, c, and d are real numbers which satisfy the system of equations
a + 2b + 3c + 4d = 10
4a + b + 2c + 3d = 4
3a + 4b + c + 2d = 10
2a + 3b + 4c + d = −4.
Find a + b + c + d.

 

\(\begin{array}{|lrcll|} \hline (1) & a + 2b + 3c + 4d &=& 10 \\ (2) & 4a + b + 2c + 3d &=& 4 \\ (3) & 3a + 4b + c + 2d &=& 10 \\ (4) &2a + 3b + 4c + d &=& −4 \\ \hline \hline \text{sum} & 10a+10b+10c+10d &=& 10 + 4 + 10 - 4 \\ & 10(a+b+c+d) &=& 20 \quad | \quad : 10 \\ & a+b+c+d &=& \dfrac{20}{10} \\ & \mathbf{a+b+c+d} &=& \mathbf{2} \\ \hline \end{array} \)

 

laugh

 Oct 26, 2020
 #4
avatar+111600 
+2

It is very nice to see you back again Heureka   wink

Melody  Oct 26, 2020

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