1057@shubhodip is the question correct bcoz
equation 1 - equation 3 gives GIF(z)-GIF(x)=2.2
how can dis be possible since both GIF(x) , GIF(z) are integers...
Find the solutions to the system of equations in reals:
x+\left \lfloor y \right \rfloor+\left \left \{ z\left. \right \} \right.=1.1
\left \lfloor x \right \rfloor+\left \{ y\left. \right \} \right.+z=2.2
\left \{ x\left. \right \} +y\right.+\left \lfloor z\right \rfloor=3.3
Where for any real x, {x} denotes the fractional part of x and \left \lfloor x \right \rfloor = x - {x}
Edit:I corrected it now
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6 Answers
1057
1708\hspace{-16}\mathbf{x+\lfloor y \rfloor+\left\{z\right\}=1.1..................(1) }$\\\\ $\mathbf{\lfloor x \rfloor+\left\{y\right\}+z=2.2..................(2)}$\\\\ $\mathbf{\left\{x\right\}+y+\lfloor z \rfloor=3.3..................(2)}$\\\\ Add equation $\mathbf{(1)+(2)+(3)}\;,$ We Get\\\\ $\mathbf{x+\underbrace{\lfloor x \rfloor +\left\{x\right\}}+y+\underbrace{\lfloor y \rfloor +\left\{y\right\}}+z+\underbrace{\lfloor z \rfloor +\left\{z\right\}}=1.1+2.2+3.3}$\\\\ $\mathbf{2x+2y+2z=6.6}$\\\\ $\mathbf{x+y+z=3.3.......................(4)}$\\\\ Now Substrace $\mathbf{(4)-(1)}$\\\\ $\mathbf{\left\{y\right\}+\lfloor z \rfloor =2.2}$\\\\ Means $\mathbf{\left\{y\right\}=0.2\;\;, \lfloor z \rfloor=2}$\\\\ Similarly Substrace $\mathbf{(4)-(2)}$\\\\ $\mathbf{\left\{x\right\}+\lfloor y \rfloor = 1.1}$\\\\ Means $\mathbf{\left\{x\right\}=0.1\;\;, \lfloor y \rfloor=1}$\\\\ Similarly Substrace $\mathbf{(4)-(3)}$\\\\ $\mathbf{\left\{z\right\}+\lfloor x \rfloor = 0.0}$\\\\
\hspace{-16}$Means $\mathbf{\left\{z\right\}=0.0\;\;, \lfloor x \rfloor=0}$\\\\\\ So \begin{Bmatrix} \bold{x =\lfloor x \rfloor +\left\{x\right\}= 0+0.1=0.1} \\\\ \bold{y =\lfloor y \rfloor +\left\{y\right\}=1+0.2=1.2} \\ \\ \bold{z =\lfloor z \rfloor +\left\{z\right\}=2+0.0=2.0} \end{Bmatrix}
1708\hspace{-16}$Solve System of equations for real $\mathbf{x\;,y\;,z}$\\\\ $\mathbf{\left\{x\right\}+\lfloor y \rfloor +\left\{z\right\}=2.9}$\\\\ $\mathbf{\left\{y\right\}+\lfloor z \rfloor +\left\{x\right\}=5.5}$\\\\ $\mathbf{\left\{z\right\}+\lfloor x \rfloor +\left\{y\right\}=4.0}$
21Solution to Man111 Problem above
By using 3rd eqn.,
{y}+{z}=1 or 0
Case 1: {y}=1- {z}
Putting in the 1st eqn. we can say that {y} - {x}=0.1
and From the 2nd eq. we get {x} + {y}=0.5 or 1.5
So, {x}=0.7 or 0.2 and {y}=0.8 or 0.3 {z}=0.2 or 0.7
x=3.7 or 3.2 y=2.8 or 2.3 and z=4.2 or 5.7
Case 2:{y}={z}=0
Putting these in the above eqns. we get no solutions.
Edit:I made a silly mistake. Thank you man111
1708Arnab I have a answer Which is
\hspace{-16}$\mathbf{$\left(x\;,y\;,z\right)=\left(3.2\;,2.3\;,5.7\right)}$\\\\ and $\mathbf{$\left(x\;,y\;,z\right)=\left(3.7\;,2.8\;,4.2\right)}$
