Question

greatest integer function

Answer 1

\(\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}}\)

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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 \(x=-3\)

Answer 9

how did u find that only \(x=-3\) satisfies this equation

\(\Large x+4+\lfloor{\dfrac{2x+1}{5}\rfloor}=0\)

Answer 10

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

Answer 11

thnks

Answer 12

yw

Answer 13

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:

Answer 14

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

Answer 15

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