# Thread: solve the equation floor

1. ## solve the equation floor

solve the equation

$\displaystyle (19x+16)/10=[(4x+7)/3]$

2. $\displaystyle (19x + 16)/10 - (4x+7)/3 = 0$

$\displaystyle (57x + 48)/30 - (40x+70)/30 = 0$

$\displaystyle ((17x - 22)/30)*30 = 0*30$

$\displaystyle 17x - 22 = 0$

$\displaystyle (17x)/17 = 22/17$

$\displaystyle x = 22/17$

3. millerst, the RHS is $\displaystyle \left\lfloor\frac{4x+7}3\right\rfloor,$ not $\displaystyle \frac{4x+7}3.$

Originally Posted by perash
solve the equation

$\displaystyle (19x+16)/10=[(4x+7)/3]$
Let $\displaystyle \left\lfloor\frac{4x+7}3\right\rfloor=n.$

Then $\displaystyle n\ \leqslant\ \frac{4x+7}3\ <\ n+1$

$\displaystyle \Rightarrow\ 3n\ \leqslant\ 4x+7\ <\ 3n+3$

$\displaystyle \Rightarrow\ \frac{3n-7}4\ \leqslant\ x\ <\ \frac{3n-4}4$

$\displaystyle \Rightarrow\ \frac{57n-69}4\ \leqslant\ 19x+16\ <\ \frac{57n-12}4$

$\displaystyle \Rightarrow\ \frac{57n-69}{40}\ \leqslant\ \frac{19x+16}{10}\ <\ \frac{57n-12}{40}$

$\displaystyle \Rightarrow\ \frac{57n-69}{40}\ \leqslant\ n\ <\ \frac{57n-12}{40}$

$\displaystyle \Rightarrow\ 57n-69\ \leqslant\ 40n\ <\ 57n-12$

$\displaystyle \Rightarrow\ -69\ \leqslant\ -17n\ <\ -12$

$\displaystyle \Rightarrow\ \frac{12}{17}\ <\ n\ \leqslant\ \frac{69}{17}$

Hence $\displaystyle n=1,2,3,4.$ The four solutions are found by solving $\displaystyle \frac{19x+16}{10}=1,2,3,4$ for $\displaystyle x.$

4. Originally Posted by perash
solve the equation

$\displaystyle (19x+16)/10=[(4x+7)/3]$
I take it from the title that this is really supposed to be
$\displaystyle (19x+ 16)/10= \lfloor (4x+7)/3\rfloor$.

The first thing that tells us is that right hand side is an integer and so (19x+ 16)/10 must be an integer: (19x+ 16)/10= n so 19x+ 16= 10n for some integer n. From that 19x= 10n-16 and x= (10n- 16)/19.

Putting that into the right side, (4x+ 7)/3= (40n+ 69)/57 and that must have "floor" n. That is, it must be equal to n+ $\displaystyle \delta$ where $\displaystyle \delta$ is between 0 and 1.

(40n+ 69)/57= n+ $\displaystyle \delta$ so $\displaystyle 40n+ 69= 57n+ 5\delta$ or $\displaystyle 17n= 69- 57\delta$ and, finally, $\displaystyle n= \frac[99- 57\delta}{17}$.

Since the largest $\displaystyle \delta$ can be is 1, n cannot be smaller than (69-57)/17= 12/17= 0.705 .... Since the smallest $\displaystyle \delta$ can be is 0, n cannot be larger than 69/17= 4.04. Since n must be a constant n must be 1, 2, 3, or 4.

If n= 1, then 19x+ 16= 10 and x= -6/19.

If n= 2, then 19x+ 16= 20 and x= 4/19.

If n= 3, then 19x+ 16= 30 and x= 14/19.

If n= 4, then 19x+ 16= 40 and x= 24/19.

I will leave it to you to check that all those do, in fact, satisfy the original equation.

Blast, too slow!! Ah, well, ladies first.