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# If $x$, $y$, and $z$ are real numbers for which? Then what is $xyz$

0
56
4

\begin{align*} x+y-z &= -8, \\ x-y+z &= 18,\text{ and} \\ -x+y+z &= 30, \\ \end{align*}

Nov 30, 2020

#1
-1

The solution is x= 5, y = 12, z = 25, so xyz = 1500.

Nov 30, 2020
#3
0

- x + y + z = 30

- 5 + 12 + 25 = 32

Guest Nov 30, 2020
#2
+1

Solve the following system:
{x + y - z = -8 | (equation 1)
x - y + z = 18 | (equation 2)
-x + y + z = 30 | (equation 3)

Subtract equation 1 from equation 2:
{x + y - z = -8 | (equation 1)
0 x - 2 y + 2 z = 26 | (equation 2)
-x + y + z = 30 | (equation 3)

Divide equation 2 by 2:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
-x + y + z = 30 | (equation 3)

Add equation 1 to equation 3:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
0 x+2 y+0 z = 22 | (equation 3)

Divide equation 3 by 2:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
0 x+y+0 z = 11 | (equation 3)

Add equation 2 to equation 3:
{x + y - z = -8 | (equation 1)
0 x - y + z = 13 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

Subtract equation 3 from equation 2:
{x + y - z = -8 | (equation 1)
0 x - y+0 z = -11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

Multiply equation 2 by -1:
{x + y - z = -8 | (equation 1)
0 x+y+0 z = 11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

Subtract equation 2 from equation 1:
{x + 0 y - z = -19 | (equation 1)
0 x+y+0 z = 11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

Add equation 3 to equation 1:
{x+0 y+0 z = 5 | (equation 1)
0 x+y+0 z = 11 | (equation 2)
0 x+0 y+z = 24 | (equation 3)

x = 5,     y = 11,      z = 24          xyz =5  x  11  x  24 =1,320

Nov 30, 2020
#4
+893
0

Good job, Guest!

jugoslav  Nov 30, 2020