greatest integer function

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- mathmath333

greatest integer function

- jamiebookeater

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- mathmath333

\(\large \color{black}{\begin{align} &\{x,y\}\in \mathbb{R} \hspace{.33em}\\~\\
& \lfloor{4x+5\rfloor}=5y+3 \hspace{.33em}\\~\\
& \lfloor{3y+7\rfloor}=x+4 \hspace{.33em}\\~\\
& \normalsize \text{where }\ \lfloor{\ \rfloor}\ \text{represents the greatest integer function } \hspace{.33em}\\~\\
&\normalsize \text{Find}\ x^2y^2 \hspace{.33em}\\~\\
& a.)\ 1 \hspace{.33em}\\~\\
& b.)\ 2 \hspace{.33em}\\~\\
& c.)\ 4 \hspace{.33em}\\~\\
& d.)\ \normalsize \text{None of these} \hspace{.33em}\\~\\
\end{align}}\)

- anonymous

I think it's an interesting system, so far I got one solution hopefully it's right. I stopped cause I believed there's no other solution. However there could be another soltion..
First start by assuming that we have either integer values orn non integers, each one will lead to a form of equation. The bonding ever however is hard to determine initially.
Assume they are integers and solve the system by any method, x=-3 and y=-2 will he satisfied

- anonymous

Since the question asks for unspecified x,y. Then it somehow gives a clue that there's only one x and one y that are satisfied. hence, the answer I THINK is none of the above

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## More answers

- mathmath333

what if there exists another solution other than \(x=-3\) and \(y=-2\)

- anonymous

Then you will in troubke. Because in order to get red of the greatest integer, then you will have to find c such that : [f(x)]=f(x)-c is an integer. This c will add a third and fourth parameter for the system.
I told you I'm not saying it's impossible, but I need to think about it

- anonymous

Another way to do it: You could try plotting both "lines". It might be a bit tedious, but it can be done. Doing so reveals two solutions, \(x=-3,y=-2\) and \(x=-2,y=-\dfrac{6}{5}\).
There's probably a much better way to go about it.

- anonymous

The lines plotted does not represent the functions we are dealing with, except if the expression inside it is integer value. Considering this case, then we ensure that we a have system of linear equations in two variables, which in turn satisfied at one point, zero points or infinite points, but never at two points ...
So necessarily the pair (-3,-2) is UNIQUE
Whenever there's another solution, then the functions we have are not linear anymore, they will be something else, in our case least integer function which are satisfied at non integer parameters

- anonymous

system:\[\lfloor{4x+5\rfloor}=5y+3 \\ \lfloor{3y+7\rfloor}=x+4\]second equation tells us that \(x+4\) is an integer, so \(4x+5\) is an integer too. It follows that\[\lfloor{4x+5\rfloor}=4x+5=5y+3 \\ y=\frac{4x+2}{5}\]put this in the second equation\[\lfloor{3\frac{4x+2}{5}+7\rfloor}=x+4 \\ 2x+8 +\lfloor{\frac{2x+1}{5}\rfloor}=x+4 \\ x+4+\lfloor{\frac{2x+1}{5}\rfloor}=0\]only integer that satisfies last equation is \(x=-3\)

- mathmath333

how did u find that only \(x=-3\) satisfies this equation
\(\Large x+4+\lfloor{\dfrac{2x+1}{5}\rfloor}=0\)

- anonymous

for \(x>-3\) LHS becomes a positive number, for \(x<-3\) LHS is negative, only possibility is \(x=-3\)

- mathmath333

thnks

- anonymous

yw

- anonymous

Both \(x\) and \(y\) are said to be real numbers, not just integers. Besides, the case where \(y=-\dfrac{6}{5}\) still works because \(5y+3\) is an integer.
Plot for the curious:

##### 1 Attachment

- anonymous

but \(\left(-2, -\frac{6}{5} \right)\) fails to satisfy second equation

- anonymous

Oh you're right, the RHS and LHS are off by one.

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