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.
$\begin{cases}a+2b+3c+4d=10\\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$
$\begin{cases}a+2b+3c+4d=10 \ \implies \ a+2b+3c+4d-\left(2b+3c+4d\right)=10-\left(2b+3c+4d\right) \ \implies \ a=10-2b-3c-4d \\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$
$\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 :\
$\begin{cases}a+2b+3c+4d=10\\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$
$\begin{cases}a+2b+3c+4d=10 \ \implies \ a+2b+3c+4d-\left(2b+3c+4d\right)=10-\left(2b+3c+4d\right) \ \implies \ a=10-2b-3c-4d \\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$
$\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 :\
Alternatively, you could just add them all up to get
10a + 10b + 10c + 10d = 0
so: a + b + c + d = 0
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!