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# help algebra

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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

#1
+164
+5

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

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

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

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

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

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

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

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
+164
+5

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

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

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

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

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

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

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

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 :\

#2
+123511
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.....

CPhill  Jul 5, 2021
#3
+32993
+3

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
+111
+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