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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.

 Jul 5, 2021

Best Answer 

 #1
avatar+139 
+2

$\begin{cases}a+2b+3c+4d=10\\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$

 

$\begin{cases}a+2b+3c+4d=10 \ \implies \ a+2b+3c+4d-\left(2b+3c+4d\right)=10-\left(2b+3c+4d\right) \ \implies \ a=10-2b-3c-4d \\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$


$\begin{cases}4\left(10-2b-3c-4d\right)+b+2c+3d=4\\ 3\left(10-2b-3c-4d\right)+4b+c+2d=-10\\ 2\left(10-2b-3c-4d\right)+3b+4c+d=-4\end{cases}  $

 

$\begin{cases}40-8b-12c-16d+b+2c+3d=4 \\ 30-6b-9c-12d+4b+c+2d=-10\\ 20-4b-6c-8d+3b+4c+d \end{cases}  $

 

$ \begin{cases}-7b-10c-13d+40=4\\ -2b-8c-10d+30=-10\\ -b-2c-7d+20=-4\end{cases}  $


$ \begin{cases}-7b-10c-13d+40=4 \ \implies \ \ -7b-10c-13d+40-\left(-10c-13d\right)=4-\left(-10c-13d\right) \ \implies \ \ -7b+40=4+10c+13d \ \implies \ -7b=10c+13d-36 \ \implies \ b=-\frac{10c+13d-36}{7}  \\ -2b-8c-10d+30=-10 \\ -b-2c-7d+20=-4\end{cases}  $

 

$ \begin{cases}-2\left(-\frac{10c+13d-36}{7}\right)-8c-10d+30=-10\\ -\left(-\frac{10c+13d-36}{7}\right)-2c-7d+20=-4\end{cases}  $

there's quite a lot of steps, used wolfram and got:

 

 $ \begin{cases}\frac{-36c-44d-72}{7}+30=-10 \implies \frac{-36c-44d-72}{7}=-40 \implies -36c-44d-72=-280 \implies -36c-72=-280+44d \implies -36c=44d-208 \implies  c=-\frac{11d-52}{9}  \\ \frac{-4c-36d-36}{7}+20=-4\end{cases}  $ 

thus, 


  $ \begin{cases}\frac{-4\left(-\frac{11d-52}{9}\right)-36d-36}{7}+20=-4\end{cases} $

$ \Updownarrow  $

$ -\frac{4\left(10d+19\right)}{9}=-24  $

$ -4\left(10d+19\right)=-216  $

$ 10d+19=54  $

$10d=35$

$ d=\frac{7}{2}  $


since we know back, lets walk all the way back for, c, then b, then a:

 

$ c=-\frac{11d-52}{9} \Leftrightarrow c=-\frac{11\cdot \frac{7}{2}-52}{9} \implies c=-\frac{\frac{77}{2}-52}{9} \implies c=-\frac{-\frac{27}{2}}{9} \implies c=-\left(-\frac{27}{18}\right) \text{  or just }  \boxed{c=\frac{27}{18}} \Leftrightarrow  \boxed{c=\frac{3}{2}}  $

 

$ b=-\frac{10c+13d-36}{7} \Leftrightarrow b=-\frac{10\cdot \frac{3}{2}+13\cdot \frac{7}{2}-36}{7} \implies b=-\frac{15+\frac{91}{2}-36}{7} \implies b=-\frac{\frac{91}{2}-21}{7} \implies b=-\frac{\frac{91}{2}-\frac{21\cdot \:2}{2}}{7} \implies b=-\frac{\frac{49}{2}}{7} \implies \boxed{-\frac{49}{14}} \Leftrightarrow \boxed{b=-\frac{7}{2}} $

 

$a=10-2b-3c-4d \Leftrightarrow a=10-2\left(-\frac{7}{2}\right)-3\cdot \frac{3}{2}-4\cdot \frac{7}{2} \leftrightarrow a=2\cdot \frac{7}{2}-4\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a=-2\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a= -\frac{14}{-2}- \frac{9}{2}+10 \implies a= -7-\frac{9}{2}+10 \implies a= -\frac{9}{2}+3 \implies a= \frac{3\cdot \:2}{2}-\frac{9}{2} \implies \boxed{\frac{-3}{2}}  $

 

thus, we got:

 

$ d=\frac{7}{2} \ \ \ ; \ \ \  b=-\frac{7}{2}  \ \ \ ; \ \ \ c=\frac{3}{2}  \ \ \ ; \ \ \   a=-\frac{3}{2}  $

 

to complete our final condition of $a+b+c+d$:

 

$ a+b+c+d= \frac{7}{2}+\left(-\frac{7}{2}\right)+\frac{3}{2}+\left(-\frac{3}{2}\right)  $

$ \boxed{a+b+c+d =0}  $

 

maybe i have messed anything up in the way :\

 Jul 5, 2021
 #1
avatar+139 
+2
Best Answer

$\begin{cases}a+2b+3c+4d=10\\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$

 

$\begin{cases}a+2b+3c+4d=10 \ \implies \ a+2b+3c+4d-\left(2b+3c+4d\right)=10-\left(2b+3c+4d\right) \ \implies \ a=10-2b-3c-4d \\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$


$\begin{cases}4\left(10-2b-3c-4d\right)+b+2c+3d=4\\ 3\left(10-2b-3c-4d\right)+4b+c+2d=-10\\ 2\left(10-2b-3c-4d\right)+3b+4c+d=-4\end{cases}  $

 

$\begin{cases}40-8b-12c-16d+b+2c+3d=4 \\ 30-6b-9c-12d+4b+c+2d=-10\\ 20-4b-6c-8d+3b+4c+d \end{cases}  $

 

$ \begin{cases}-7b-10c-13d+40=4\\ -2b-8c-10d+30=-10\\ -b-2c-7d+20=-4\end{cases}  $


$ \begin{cases}-7b-10c-13d+40=4 \ \implies \ \ -7b-10c-13d+40-\left(-10c-13d\right)=4-\left(-10c-13d\right) \ \implies \ \ -7b+40=4+10c+13d \ \implies \ -7b=10c+13d-36 \ \implies \ b=-\frac{10c+13d-36}{7}  \\ -2b-8c-10d+30=-10 \\ -b-2c-7d+20=-4\end{cases}  $

 

$ \begin{cases}-2\left(-\frac{10c+13d-36}{7}\right)-8c-10d+30=-10\\ -\left(-\frac{10c+13d-36}{7}\right)-2c-7d+20=-4\end{cases}  $

there's quite a lot of steps, used wolfram and got:

 

 $ \begin{cases}\frac{-36c-44d-72}{7}+30=-10 \implies \frac{-36c-44d-72}{7}=-40 \implies -36c-44d-72=-280 \implies -36c-72=-280+44d \implies -36c=44d-208 \implies  c=-\frac{11d-52}{9}  \\ \frac{-4c-36d-36}{7}+20=-4\end{cases}  $ 

thus, 


  $ \begin{cases}\frac{-4\left(-\frac{11d-52}{9}\right)-36d-36}{7}+20=-4\end{cases} $

$ \Updownarrow  $

$ -\frac{4\left(10d+19\right)}{9}=-24  $

$ -4\left(10d+19\right)=-216  $

$ 10d+19=54  $

$10d=35$

$ d=\frac{7}{2}  $


since we know back, lets walk all the way back for, c, then b, then a:

 

$ c=-\frac{11d-52}{9} \Leftrightarrow c=-\frac{11\cdot \frac{7}{2}-52}{9} \implies c=-\frac{\frac{77}{2}-52}{9} \implies c=-\frac{-\frac{27}{2}}{9} \implies c=-\left(-\frac{27}{18}\right) \text{  or just }  \boxed{c=\frac{27}{18}} \Leftrightarrow  \boxed{c=\frac{3}{2}}  $

 

$ b=-\frac{10c+13d-36}{7} \Leftrightarrow b=-\frac{10\cdot \frac{3}{2}+13\cdot \frac{7}{2}-36}{7} \implies b=-\frac{15+\frac{91}{2}-36}{7} \implies b=-\frac{\frac{91}{2}-21}{7} \implies b=-\frac{\frac{91}{2}-\frac{21\cdot \:2}{2}}{7} \implies b=-\frac{\frac{49}{2}}{7} \implies \boxed{-\frac{49}{14}} \Leftrightarrow \boxed{b=-\frac{7}{2}} $

 

$a=10-2b-3c-4d \Leftrightarrow a=10-2\left(-\frac{7}{2}\right)-3\cdot \frac{3}{2}-4\cdot \frac{7}{2} \leftrightarrow a=2\cdot \frac{7}{2}-4\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a=-2\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a= -\frac{14}{-2}- \frac{9}{2}+10 \implies a= -7-\frac{9}{2}+10 \implies a= -\frac{9}{2}+3 \implies a= \frac{3\cdot \:2}{2}-\frac{9}{2} \implies \boxed{\frac{-3}{2}}  $

 

thus, we got:

 

$ d=\frac{7}{2} \ \ \ ; \ \ \  b=-\frac{7}{2}  \ \ \ ; \ \ \ c=\frac{3}{2}  \ \ \ ; \ \ \   a=-\frac{3}{2}  $

 

to complete our final condition of $a+b+c+d$:

 

$ a+b+c+d= \frac{7}{2}+\left(-\frac{7}{2}\right)+\frac{3}{2}+\left(-\frac{3}{2}\right)  $

$ \boxed{a+b+c+d =0}  $

 

maybe i have messed anything up in the way :\

UsernameTooShort Jul 5, 2021
 #2
avatar+120966 
0

Excellent   job, UNTS  !!!!!    (and  the answer IS  correct......   )

 

I'd  give you  five points  (if I could)  for  all that  effort   !!!

 

Alas......you  will just  have to settle for  "Best Answer"  and 1 point from me.....sad

 

 

cool cool cool

CPhill  Jul 5, 2021
 #3
avatar+32399 
+2

Alternatively, you could just add them all up to get

 

10a + 10b + 10c + 10d = 0

 

so:    a + b + c + d = 0

 Jul 5, 2021
 #4
avatar+114 
+1

LMFAO!!

Hours of complex, laborious work and a few seconds of WolframAlpha computer time, all to add 4 zeros. 

This would make great cannon fodder for one of GingerAle’s troll posts!

 Jul 6, 2021

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