Solve in $\mathbb{R}$ : $2x+5y+4u+3v=10$, $3x+3y+2u+2v=6$, $x+7y+8u+2v=-6$, $3x+5y+6u+2v=-10$

Question

Solve in $\mathbb{R}$ the system equation following :

$$\begin{cases}2x+5y+4u+3v=10\\3x+3y+2u+2v=6\\x+7y+8u+2v=-6\\3x+5y+6u+2v=-10\end{cases}$$

I tried many ways like get $x$ from equation $(1)$ and complete but I didn't find any things but after adding equation $(1)$ and $(4)$,$(2)$ and $(3)$ I get :

$$\begin{cases}5x+5v+10y+10u=0\\4x+4v+10y+10u=0\end{cases}$$

From here I think take substation $t=x+u$ and $z=y+v$ I get :

$$\begin{cases}\\5t+10z=0\\4t+10z=0\end{cases}$$ Its clearly that $z=t=0$

Then how I can complete ??

Answer 1

HINT

So you proved $u = -x$ and $v = -y$. Can you plug into the original system, get 2 equations and solve?

BTW usually one eliminates each variable one-by-one to avoid the problem, the procedure is systematized in the Gaussian Elimination algorithm.

Answer 2

$(2)-(4): \ -2y-4u=16$

and

$(1)-2(3): \ -9y-12u-v=24$

and

$(2)-3(3):\ -18y-22u-4v=24$

so we have that the initial $\Rightarrow \begin{cases} -2y-4u=16\\ -9y-12u-v=24\\ 18y-22u-4v=24\end{cases}$, can you solve this?