solve the equation floor

perash · #1 $Old$ November 9th, 2008, 01:40 PM

solve the equation

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

millerst · #2 $Old$ November 9th, 2008, 04:12 PM

$(19x + 16)/10 - (4x+7)/3 = 0$

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

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

$17x - 22 = 0$

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

$x = 22/17$

JaneBennet · #3 $Old$ November 10th, 2008, 05:07 AM

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

Quote:

Originally Posted by perash $View Post$

solve the equation

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

Let $\left\lfloor\frac{4x+7}3\right\rfloor=n.$

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

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

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

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

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

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

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

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

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

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

HallsofIvy · #4 $Old$ November 10th, 2008, 05:38 AM

Quote:

Originally Posted by perash $View Post$

solve the equation

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

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

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+ $\delta$ where $\delta$ is between 0 and 1.

(40n+ 69)/57= n+ $\delta$ so $40n+ 69= 57n+ 5\delta$ or $17n= 69- 57\delta$ and, finally, [LaTeX Error: Syntax error].

Since the largest $\delta$ can be is 1, n cannot be smaller than (69-57)/17= 12/17= 0.705 .... Since the smallest $\delta$ 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.