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resoudre-dans-R-3-x-y-3-y-z-5-x-z-4-




Question Number 208420 by lepuissantcedricjunior last updated on 15/Jun/24
resoudre dans R^3    { ((x+y=3)),((y+z=5)) :}x+z=4
$$\boldsymbol{{resoudre}}\:\boldsymbol{{dans}}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{3}}\\{\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{5}}\end{cases}\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{4} \\ $$
Commented by A5T last updated on 15/Jun/24
You should learn to signify that you changed  a question.  This was your original question:   { ((x+y=3)),((y−z=−1)) :}x+z=4   ((1,1,0),(1,0,1),(0,1,(−1)) ) ((x),(y),(z) )= ((3),(4),((−1)) )
$${You}\:{should}\:{learn}\:{to}\:{signify}\:{that}\:{you}\:{changed} \\ $$$${a}\:{question}. \\ $$$${This}\:{was}\:{your}\:{original}\:{question}: \\ $$$$\begin{cases}{\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{3}}\\{\boldsymbol{{y}}−\boldsymbol{{z}}=−\mathrm{1}}\end{cases}\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{4} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}&{−\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{4}}\\{−\mathrm{1}}\end{pmatrix} \\ $$
Answered by A5T last updated on 15/Jun/24
(x,3−x,4−x) infinitely many solutions
$$\left({x},\mathrm{3}−{x},\mathrm{4}−{x}\right)\:{infinitely}\:{many}\:{solutions} \\ $$
Commented by A5T last updated on 15/Jun/24
Even one equation can have infinitely many   solutions : x+y=1
$${Even}\:{one}\:{equation}\:{can}\:{have}\:{infinitely}\:{many}\: \\ $$$${solutions}\::\:{x}+{y}=\mathrm{1} \\ $$
Commented by lepuissantcedricjunior last updated on 15/Jun/24
no!!!  c′est trois equations
$${no}!!! \\ $$$$\boldsymbol{{c}}'{est}\:\boldsymbol{{trois}}\:\boldsymbol{{equations}} \\ $$
Answered by Frix last updated on 15/Jun/24
1+2=3  1+3=4  2+3=5  x=1 y=2 z=3
$$\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\mathrm{1}+\mathrm{3}=\mathrm{4} \\ $$$$\mathrm{2}+\mathrm{3}=\mathrm{5} \\ $$$${x}=\mathrm{1}\:{y}=\mathrm{2}\:{z}=\mathrm{3} \\ $$

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