An easy one...
Given that x , y and z are non-zero real numbers such that
x + y = 5 x y ;
y + z = 6 y z ;
x + z = 7 x z ,
find the value of x + y + z rounded to 2 decimal places.
The answer is 1.08.
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2 solutions
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In order to divide by x, y, z, you need the condition that x y z = 0 . I've added in the condition that x, y, z are non-zero real numbers.
Note that x = y = z = 0 is also a solution, which would yield x + y + z = 0 . Those who previously answered 0 have been marked correct.
As
x
and
y
are nonzero real numbers we can divide both sides of the first equation
x
+
y
=
5
x
y
by
x
y
, then you get the equivalent equation
x
1
+
y
1
=
5
(
1
)
In a similar way, using the second and third given equations, respectively, we get
z
1
+
y
1
=
6
(
2
)
and
z
1
+
x
1
=
7
(
3
)
Adding the equations (1), (2), and (3) and dividing both sides of the resulting equation by 2 we get
x
1
+
y
1
+
z
1
=
9
(
4
)
Subtracting the equations (1), (2) and (3) from (4), respectively, we obtain that
z
1
=
4
,
x
1
=
3
, and
y
1
=
2
. Therefore
x
+
y
+
z
=
.
3
3
3
.
.
.
+
0
.
5
+
0
.
2
5
Rounding this result to 2 decimal places you get 1.08.
(X+Y)/XY - (Y+Z)/YZ - (Z+X)/ZX =5-6-7.
On simplification we get
-2XY=-8XYZ
Z=1/4
Using the value of Z in 2nd & 3rd equation we get
Y=1/2 and X=1/3
So X+Y+Z = 1.0833