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.
+175 ${a+2b+3c+4d=104a+b+2c+3d=43a+4b+c+2d=−102a+3b+4c+d=−4$
${a+2b+3c+4d=10 ⟹ a+2b+3c+4d−(2b+3c+4d)=10−(2b+3c+4d) ⟹ a=10−2b−3c−4d4a+b+2c+3d=43a+4b+c+2d=−102a+3b+4c+d=−4$
${4(10−2b−3c−4d)+b+2c+3d=43(10−2b−3c−4d)+4b+c+2d=−102(10−2b−3c−4d)+3b+4c+d=−4 $
${40−8b−12c−16d+b+2c+3d=430−6b−9c−12d+4b+c+2d=−1020−4b−6c−8d+3b+4c+d $
$ {−7b−10c−13d+40=4−2b−8c−10d+30=−10−b−2c−7d+20=−4 $
$ {−7b−10c−13d+40=4 ⟹ −7b−10c−13d+40−(−10c−13d)=4−(−10c−13d) ⟹ −7b+40=4+10c+13d ⟹ −7b=10c+13d−36 ⟹ b=−10c+13d−367−2b−8c−10d+30=−10−b−2c−7d+20=−4 $
$ {−2(−10c+13d−367)−8c−10d+30=−10−(−10c+13d−367)−2c−7d+20=−4 $
there's quite a lot of steps, used wolfram and got:
$ {−36c−44d−727+30=−10⟹−36c−44d−727=−40⟹−36c−44d−72=−280⟹−36c−72=−280+44d⟹−36c=44d−208⟹c=−11d−529−4c−36d−367+20=−4 $
thus,
$ {−4(−11d−529)−36d−367+20=−4 $
$ \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 :\
+175 ${a+2b+3c+4d=104a+b+2c+3d=43a+4b+c+2d=−102a+3b+4c+d=−4$
${a+2b+3c+4d=10 ⟹ a+2b+3c+4d−(2b+3c+4d)=10−(2b+3c+4d) ⟹ a=10−2b−3c−4d4a+b+2c+3d=43a+4b+c+2d=−102a+3b+4c+d=−4$
${4(10−2b−3c−4d)+b+2c+3d=43(10−2b−3c−4d)+4b+c+2d=−102(10−2b−3c−4d)+3b+4c+d=−4 $
${40−8b−12c−16d+b+2c+3d=430−6b−9c−12d+4b+c+2d=−1020−4b−6c−8d+3b+4c+d $
$ {−7b−10c−13d+40=4−2b−8c−10d+30=−10−b−2c−7d+20=−4 $
$ {−7b−10c−13d+40=4 ⟹ −7b−10c−13d+40−(−10c−13d)=4−(−10c−13d) ⟹ −7b+40=4+10c+13d ⟹ −7b=10c+13d−36 ⟹ b=−10c+13d−367−2b−8c−10d+30=−10−b−2c−7d+20=−4 $
$ {−2(−10c+13d−367)−8c−10d+30=−10−(−10c+13d−367)−2c−7d+20=−4 $
there's quite a lot of steps, used wolfram and got:
$ {−36c−44d−727+30=−10⟹−36c−44d−727=−40⟹−36c−44d−72=−280⟹−36c−72=−280+44d⟹−36c=44d−208⟹c=−11d−529−4c−36d−367+20=−4 $
thus,
$ {−4(−11d−529)−36d−367+20=−4 $
$ \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 :\
+33657 Alternatively, you could just add them all up to get
10a + 10b + 10c + 10d = 0
so: a + b + c + d = 0
+112 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!
