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Assuming \(x\), \(y\), and \(z\) are positive real numbers satisfying
\(xy - z = 15\\ xz - y = 15\\ yz - x = 15\)
then, what is the value of \(xyz\)?

\(\begin{array}{|lrcll|} \hline (1) & xy - z &=& 15 \\ (2) & xz - y &=& 15 \\ (3) & yz - x &=& 15 \\ \hline (1)+(2)+(3):& (xy - z)+(xz - y)+(yz - x) &=& 45 \\ & \ldots \\ &\mathbf{ x+y+z } &=& \mathbf{ xy+xz+yz-45 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline yz - x &=& 15 \\ \mathbf{x} &=& yz-15 \\ \hline xz-y &=& 15 \\ (yz-15)z-y &=& 15 \\ yz^2-15z-y &=& 15 \\ yz^2-y &=& 15z+15 \\ y(z^2-1) &=& 15(z+1) \\ y(z-1)(z+1) &=& 15(z+1) \quad | \quad : (z+1),~ z>0!\\ y(z-1)&=& 15\\ \mathbf{y} &=& \mathbf{\dfrac{15} {z-1} }\qquad (4) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline xz - y &=& 15 \\ \mathbf{y} &=& xz-15 \\ \hline xy-z &=& 15 \\ (xz-15)x-z &=& 15 \\ zx^2-15x-z &=& 15 \\ zx^2-z &=& 15x+15 \\ z(x^2-1) &=& 15(x+1) \\ z(x-1)(x+1) &=& 15(x+1) \quad | \quad : (x+1),~ x>0!\\ z(x-1)&=& 15\\ \mathbf{z} &=& \mathbf{\dfrac{15} {x-1} }\qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline xy - z &=& 15 \\ \mathbf{z} &=& xy-15 \\ \hline yz-x &=& 15 \\ (xy-15)y-x &=& 15 \\ xy^2-15y-x &=& 15 \\ xy^2-x &=& 15y+15 \\ x(y^2-1) &=& 15(y+1) \\ x(y-1)(y+1) &=& 15(y+1) \quad | \quad : (y+1),~ y>0!\\ x(y-1)&=& 15\\ \mathbf{x} &=& \mathbf{\dfrac{15} {y-1} }\qquad (6) \\ \hline \end{array}\)

\( \small{ \begin{array}{|lrcll|} \hline (4)\times(5)\times(6):& xyz &=& \dfrac{15} {y-1} \times \dfrac{15} {z-1} \times \dfrac{15} {x-1} \\\\ & xyz &=& \dfrac{15^3} {(x-1)(y-1)(z-1)} \\\\ &&& \boxed{\mathbf{(x-1)(y-1)(z-1)}\\=xyz-(xy+xz+yz)+(x+y+z) - 1} \\\\ & xyz &=& \dfrac{15^3} {xyz-(xy+xz+yz)+(x+y+z) - 1} \\\\ &&& \mathbf{ x+y+z=xy+xz+yz-45 } \\\\ & xyz &=& \dfrac{15^3} {xyz-(xy+xz+yz)+(xy+xz+yz-45) - 1} \\\\ & xyz &=& \dfrac{15^3} {xyz-46} \\\\ & xyz(xyz-46) &=& 15^3 \\\\ & (xyz)^2-46xyz- 15^3 &=& 0 \\ \hline & xyz &=& \dfrac{46\pm\sqrt{46^2-4*(-15^3)}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{46^2+4*15^3}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{15616}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{16^2*61}}{2} \\\\ & xyz &=& \dfrac{46\pm 16\sqrt{61}}{2} \\\\ & \mathbf{ xyz } &=& \mathbf{ 23 +8\sqrt{61} } \qquad xyz >0! \\ \hline \end{array} }\)

Point D is on side AC of triangle ABC such that angle ADB = angle ABD and angle ABC - angle ACB = 37 degrees.  Find angle DBC in degrees.

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Let triangle ABC be the right triangle.

∠B + ∠C = 90º          ∠ABD = ∠ADB = 45º

∠B = ∠C + 37          ∠C = x

x + 37 + x = 90          x = 26.5

∠DBC = ∠B - ∠ABD = 18.5º

The letters A, B, C, D, E, and F represent digits and 

ABC,DEF represents a positive six-digit integer.

What is the number ABC,DEF if 4(ABC,DEF) = 9(DEF,ABC)?

LMFAO!!

Hours of complex, laborious work and a few seconds of WolframAlpha computer time, all to add 4 zeros. 

This would make great cannon fodder for one of GingerAle’s troll posts!

y = 22.

The graph of the parabola is y = 1/2*x^2 - 3x + 7/2, so a*b*c = -21/4.

The largest possible value of N is 111.

Let  the  smaller amt = x        and  the larger amt =  2x

So

3x +  117  =  2 (2x)

3x  + 117   =  4x

117  =  x

The  money  will  be  divided  as     x :   2x    =           117   :   234

(x +  4)  / ( x + 5)  = ( x + 5)  / (2x)      cross-multiply

2x ( x + 4)  =   ( x + 5) ( x + 5)

2x^2  + 8x  =   x^2  + 10x  + 25       rearrange  as

x^2 - 2x    =    25                complete  the square on x

x^2  - 2x   + 1   =  25 +  1

(x - 1)^2  =   26             take  both roots

x  - 1  = ±sqrt (26)

x =   1  ±  sqrt (26)

First  series

First term  = 12

Common ratio  = 4/12  = 1/3   =  r

Sum  =  12  / ( 1 - 1/3)  =   12/ (2/3)  =   18

Second series

First term  =   16

Common  ratio   ( 2 + n)  /16

Sum  =      16   /   [ 1 - (2 + n)/16 ]    =  72

Simplify

16   /  [  (14 - n)  / 16 ]   =  72

256  / ( 14 - n)   =  72

256  =  72 ( 14  - n)

256  = 1008  -  72n

72n =  1008  - 256

72n  = 752

n =  752/72  =   94/9

Using the quadratic equation, the answers are 1 + √26 and 1 - √26

C  = 

[ 14  +  (1/2)(14 - 2)   ,   4 + (1/2) ( 4 - -2)  ]  =

[ 14  + (1/2)(12)   ,   4  + (1/2) (6)   ]  =

[ 14 + 6  ,    4  + 3   ] =

( 20 , 7 )

9x  +  7y   =   330

The slope of  this line =  - (9/7)

So

y = (-9/7)(x - 9)   - 3

y  = (-9/7)x  + 81/7  - 3

y =  (-9/7)x  + 81/7  - 21/7

y = (-9/7) x  +  60/7      

7y   =  -9x  + 60

9x + 7y   =   60

Angle AEG  = angle EGF  =  x

And  because  EF  = FG , then angles  EGF  and GEF  are  equal  =  x

So

angle   BEF  +  angle GEF  +  angle  AEG  = 180

So

130 + x   + x  +  x   =  180

3x  + 130   =  180

3x   =180  -130

3x  =   50

x  = (50/3)°   =   (16 + 2/3)°

-4.2t +  42t  + 18.9

Time  to reach  max  ht =      -42  /  (2 * -4.2)   =  -42 / -8.4  =  5  sec

x    +     x+ d      + x + 2d     + x + 3d    = 360          x + 3d = 128     or    x = 128 - 3d    <======sub this into the equation

4 ( 128-3d) + 6d = 360

d = 25 1/3

x + 3d = 128

x = 128 -3d = 128 - 3 ( 25 1/3) = 52 degrees

Alternatively, you could just add them all up to get

10a + 10b + 10c + 10d = 0

so:    a + b + c + d = 0

See answer at https://web2.0calc.com/questions/this-problem-hurts#r2

As follows:

Excellent   job, UNTS  !!!!!    (and  the answer IS  correct......   )

I'd  give you  five points  (if I could)  for  all that  effort   !!!

Alas......you  will just  have to settle for  "Best Answer"  and 1 point from me.....

Simplifying gives x^55/x^110, which equals 1/x^55. The answer should be 1/(2^55).

$\begin{cases}a+2b+3c+4d=10\\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$

$\begin{cases}a+2b+3c+4d=10 \ \implies \ a+2b+3c+4d-\left(2b+3c+4d\right)=10-\left(2b+3c+4d\right) \ \implies \ a=10-2b-3c-4d \\ 4a+b+2c+3d=4\\ 3a+4b+c+2d=-10\\ 2a+3b+4c+d=-4\end{cases}$

$\begin{cases}4\left(10-2b-3c-4d\right)+b+2c+3d=4\\ 3\left(10-2b-3c-4d\right)+4b+c+2d=-10\\ 2\left(10-2b-3c-4d\right)+3b+4c+d=-4\end{cases}  $

$\begin{cases}40-8b-12c-16d+b+2c+3d=4 \\ 30-6b-9c-12d+4b+c+2d=-10\\ 20-4b-6c-8d+3b+4c+d \end{cases}  $

$ \begin{cases}-7b-10c-13d+40=4\\ -2b-8c-10d+30=-10\\ -b-2c-7d+20=-4\end{cases}  $

$ \begin{cases}-7b-10c-13d+40=4 \ \implies \ \ -7b-10c-13d+40-\left(-10c-13d\right)=4-\left(-10c-13d\right) \ \implies \ \ -7b+40=4+10c+13d \ \implies \ -7b=10c+13d-36 \ \implies \ b=-\frac{10c+13d-36}{7}  \\ -2b-8c-10d+30=-10 \\ -b-2c-7d+20=-4\end{cases}  $

$ \begin{cases}-2\left(-\frac{10c+13d-36}{7}\right)-8c-10d+30=-10\\ -\left(-\frac{10c+13d-36}{7}\right)-2c-7d+20=-4\end{cases}  $

there's quite a lot of steps, used wolfram and got:

 $ \begin{cases}\frac{-36c-44d-72}{7}+30=-10 \implies \frac{-36c-44d-72}{7}=-40 \implies -36c-44d-72=-280 \implies -36c-72=-280+44d \implies -36c=44d-208 \implies  c=-\frac{11d-52}{9}  \\ \frac{-4c-36d-36}{7}+20=-4\end{cases}  $ 

thus, 

  $ \begin{cases}\frac{-4\left(-\frac{11d-52}{9}\right)-36d-36}{7}+20=-4\end{cases} $

$ \Updownarrow  $

$ -\frac{4\left(10d+19\right)}{9}=-24  $

$ -4\left(10d+19\right)=-216  $

$ 10d+19=54  $

$10d=35$

$ d=\frac{7}{2}  $

since we know back, lets walk all the way back for, c, then b, then a:

$ c=-\frac{11d-52}{9} \Leftrightarrow c=-\frac{11\cdot \frac{7}{2}-52}{9} \implies c=-\frac{\frac{77}{2}-52}{9} \implies c=-\frac{-\frac{27}{2}}{9} \implies c=-\left(-\frac{27}{18}\right) \text{  or just }  \boxed{c=\frac{27}{18}} \Leftrightarrow  \boxed{c=\frac{3}{2}}  $

$ b=-\frac{10c+13d-36}{7} \Leftrightarrow b=-\frac{10\cdot \frac{3}{2}+13\cdot \frac{7}{2}-36}{7} \implies b=-\frac{15+\frac{91}{2}-36}{7} \implies b=-\frac{\frac{91}{2}-21}{7} \implies b=-\frac{\frac{91}{2}-\frac{21\cdot \:2}{2}}{7} \implies b=-\frac{\frac{49}{2}}{7} \implies \boxed{-\frac{49}{14}} \Leftrightarrow \boxed{b=-\frac{7}{2}} $

$a=10-2b-3c-4d \Leftrightarrow a=10-2\left(-\frac{7}{2}\right)-3\cdot \frac{3}{2}-4\cdot \frac{7}{2} \leftrightarrow a=2\cdot \frac{7}{2}-4\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a=-2\cdot \frac{7}{2}-3\cdot \frac{3}{2}+10 \implies a= -\frac{14}{-2}- \frac{9}{2}+10 \implies a= -7-\frac{9}{2}+10 \implies a= -\frac{9}{2}+3 \implies a= \frac{3\cdot \:2}{2}-\frac{9}{2} \implies \boxed{\frac{-3}{2}}  $

thus, we got:

$ d=\frac{7}{2} \ \ \ ; \ \ \  b=-\frac{7}{2}  \ \ \ ; \ \ \ c=\frac{3}{2}  \ \ \ ; \ \ \   a=-\frac{3}{2}  $

to complete our final condition of $a+b+c+d$:

$ a+b+c+d= \frac{7}{2}+\left(-\frac{7}{2}\right)+\frac{3}{2}+\left(-\frac{3}{2}\right)  $

$ \boxed{a+b+c+d =0}  $

maybe i have messed anything up in the way :\