Question

If x+y+xy=3, y+z+yz=8, z+x+xz=15 Then find 6x

Answer 1

Here two values of 6xyz are possible.

#1,if x, y and are z positive

#2,if y, x and z are negative

Then, given equation are also written as;

x + y + xy=3 =>1 + x + y + xy =4 =>(1+ x ) (1+ y )=4

y + z + yz=8 =>1+ y + z + yz=9 => (1+ y )(1 + z )=9

Similarly,

z+ x + zx=15 =>1+ z + x + zx=16 =>(1+z )(1+x)=16

Multiplying the first and third equations and then divides by the second equation we get,

( 1 + x )²(1+ y)(1 + z)/(1+y)(1+ z) = 4×16/9 = 64/9

=>(1 + x)² = 64/9

=>1 + x = ±√(64/9)

=>1 + x =± 8/3

=> x = -1 ± 8/3

=>x= (–1 + 8/3) or (–1 – 8/3)

x = 5/3 or -11/3.

From the first equation:

x + ( 1 + x )y = 3

=>(1 + x)y = 3 – x

y = ( 3 - x ) / ( 1 + x )

From the third equation:

x + ( 1 + x ) z = 15

=>(1 + x)z =15 –x

=> z = ( 15 - x ) / ( 1 + x )

6xyz =6x.(3 –x)/(1 + x) . (15–x)/(1 + x)

= 6x ( 3 - x )/( 15 - x ) / ( 1 + x )²

= 6x ( 3 - x ) ( 15 - x ) / ( 64/9 )

= 27x ( 3 - x ) ( 15 - x ) / 32

= 3x ( 9 - 3x ) ( 45 - 3x ) / 32

Case 1; put x = 5/3 in above equation we get

6xyz = 5 × ( 9 - 5 ) × ( 45 - 5 ) / 32

= 5 × 4 × 40 / 32

= 5 × 1 × 5

6xyz = 25,Ans

Answer 2

The required answer is 30 or -42.

(*This answer is an edited one.)

 

Given equations:

$x+y+xy=3$

$y+z+yz=8$

$z+x+xz=15$

 

We can find that $(a+1)(b+1)=\underline{ab+a+b}+1$.

Using this fact, we can factorize the equations.

 

Given equations:

$(x+1)(y+1)=4$

$(y+1)(z+1)=9$

$(z+1)(x+1)=16$

 

There is a known way to solve this system equation.

We start off by multiplying altogether.

$(x+1)^2(y+1)^2(z+1)^2=24^2$

$\implies (x+1)(y+1)(z+1)=\pm24$

 

Then, we divide by each system equations.

$x+1=\pm6$

$y+1=\pm\dfrac{8}{3}$

$z+1=\pm\dfrac{3}{2}$

 

Solutions

(Note that ± sign should have the same signs.)

$\therefore x=-1\pm6$

$\therefore y=-1\pm\dfrac{8}{3}$

$\therefore z=-1\pm\dfrac{3}{2}$