UsernameTooShort

avatar
Nom d'utilisateurUsernameTooShort
But150
Membership
Stats
Questions 0
Réponses 52

 #1
avatar+150 
+3

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

 

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


$\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 :\

5 juil. 2021
 #1
avatar+150 
+2

oh lord, this is hard to swallow 

 

oh well, lets start shall we -- pay attention to every step:

 

 

$\Huge\text{NOTE:}$  this website kind of screwed up the colouring i do not know why but check this link for a more accurate version: http://mathb.in/59340

 

---------------------------------------------------------------------------------------------------------------------------------------------------

 


$ \color{lightskyblue}{\sqrt{60x}} \sqrt{12x}\sqrt{63x}\sqrt{42x} $

 

$  \color{lightskyblue}{\sqrt{4(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}  $

 

$ \color{lightskyblue}{ \sqrt{2^2(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}   $

 

$   \color{lightskyblue}{2\sqrt{(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}    $

 

now lets do the second one: $ 2\sqrt{(15)x}   \color{tomato}{\sqrt{12x}}\sqrt{63x}\sqrt{42x} $

 

$ 2\sqrt{(15)x}   \color{tomato}{ \sqrt{4(3)x}}\sqrt{63x}\sqrt{42x}  $

 

$ 2\sqrt{(15)x}  \color{tomato}{ \sqrt{2^2(3)x}}\sqrt{63x}\sqrt{42x}    $

 

$   2\sqrt{(15)x}   \color{tomato}{ 2 \sqrt{(3)x}}\sqrt{63x}\sqrt{42x}   $

 

now lets do the third one: $  2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{63x}} \sqrt{42x}     $

 

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{9(7)x}} \sqrt{42x}    $

 

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{3^3(7)x}} \sqrt{42x}    $

 

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{(3\sqrt{(7)x}) } \sqrt{42x}    $

 

We are left with

 

$ 2\sqrt{(15)x}2 \sqrt{(3)x} (3\sqrt{(7)x})  \sqrt{42x}  $ 

 

firstly lets multiply the first two terms:

 

$ \color{darkorange}{2\sqrt{(15)x}2 \sqrt{(3)x}} (3\sqrt{(7)x})  \sqrt{42x}  $ 

 

$ \color{darkorange}{4\sqrt{(15)x} \sqrt{(3)x}} (3\sqrt{(7)x})  \sqrt{42x}  $

 

$ \color{darkorange}{4\sqrt{3x(15x)}} (3\sqrt{(7)x})  \sqrt{42x}  $

 

$ \color{darkorange}{4\sqrt{45x^2}} (3\sqrt{(7)x})  \sqrt{42x}  $

 

$ \color{darkorange}{4\sqrt{9(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x}  $

 

$ \color{darkorange}{4\sqrt{3^2(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x} \Leftrightarrow \color{darkorange}{4\sqrt{3^2 \cdot x^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}  $

 

$\color{darkorange}{4\sqrt{(3x)^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}  $

 

$\color{darkorange}{4[(3x)\sqrt{5}]} (3\sqrt{(7)x}) \sqrt{42x}  $

 

$  \color{darkorange}{12x\sqrt{5}} (3\sqrt{(7)x}) \sqrt{42x}  $

 

now lets do work the $12x\sqrt{5} (3\sqrt{(7)x})$ :

 

$\color{mediumseagreen}{12x\sqrt{5} (3\sqrt{(7)x})} \sqrt{42x}$  

 

$\color{mediumseagreen}{36x\sqrt{7x\cdot 5} } \sqrt{42x}$

 

$\color{mediumseagreen}{36x\sqrt{35x} } \sqrt{42x}$

 

finally, we multiply them all together:

 

$\color{navy}{36x\sqrt{35x}  \sqrt{42x}}$

 

$\color{navy}{36x\sqrt{42x(35x)}}$

 

$\color{navy}{36x\sqrt{1470x^2}}$

 

$\color{navy}{36x\sqrt{(7x)^2 \cdot 30 }}$

 

$\color{navy}{36x(7x\sqrt{30})}$

 

$\color{navy}{36x\cdot x(7\sqrt{30})}$


$\color{navy}{36x^2(7\sqrt{30})}$

 

$\color{navy}{252x^2(\sqrt{30})}$

 

or just 

$\color{navy}{252x^2\sqrt{30}}$

 

$\color{navy}{\left(6\sqrt{7}\right)x^2\sqrt{30}}$

 

$\color{navy}{6\sqrt{7\cdot \:30}x^2}$
 
$ \boxed{\color{navy}{ 6\sqrt{210}x^2}} $ 

 

can you do anything else to this? i dont think so.
 

3 juil. 2021