Help on question

To solve the given system of equations

$$\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4,$$

we first rewrite each equation in terms of its cross-multiplication:

1. $\frac{xy}{x + y} = 1$

$$xy = x + y$$

$$xy - x - y = 0 \quad \Rightarrow \quad xy - x - y + 1 = 1$$

$$(x-1)(y-1) = 2$$

2. $\frac{xz}{x + z} = 2$

$$xz = 2(x + z)$$

$$xz - 2x - 2z = 0 \quad \Rightarrow \quad xz - 2x - 2z + 4 = 4$$

$$(x-2)(z-2) = 8$$

3. $\frac{yz}{y + z} = 4$

$$yz = 4(y + z)$$

$$yz - 4y - 4z = 0 \quad \Rightarrow \quad yz - 4y - 4z + 16 = 16$$

$$(y-4)(z-4) = 20$$

We now have three transformed equations:

$$(x-1)(y-1) = 2$$

$$(x-2)(z-2) = 8$$

$$(y-4)(z-4) = 20$$

We solve these equations step by step. Start with the first equation:

$$(x-1)(y-1) = 2$$

Express $y$ in terms of $x$:

$$y = \frac{2}{x-1} + 1$$

Next, substitute $y$ into the second equation:

$$(x-2)(z-2) = 8$$

Express $z$ in terms of $x$:

$$z = \frac{8}{x-2} + 2$$

Substitute both $y$ and $z$ into the third equation:

$$(y-4)(z-4) = 20$$

Substituting $y = \frac{2}{x-1} + 1$ and $z = \frac{8}{x-2} + 2$:

$$\left( \frac{2}{x-1} + 1 - 4 \right) \left( \frac{8}{x-2} + 2 - 4 \right) = 20$$

Simplify $y - 4$ and $z - 4$:

$$y - 4 = \frac{2}{x-1} - 3$$

$$z - 4 = \frac{8}{x-2} - 2$$

So, the equation becomes:

$$\left( \frac{2 - 3(x-1)}{x-1} \right) \left( \frac{8 - 2(x-2)}{x-2} \right) = 20$$

Simplify the terms inside the parentheses:

$$\left( \frac{2 - 3x + 3}{x-1} \right) \left( \frac{8 - 2x + 4}{x-2} \right) = 20$$

$$\left( \frac{5 - 3x}{x-1} \right) \left( \frac{12 - 2x}{x-2} \right) = 20$$

Solve for $x$. By testing rational values, we find:

Let $x = 3$:

$$y = \frac{2}{3-1} + 1 = 2$$

$$z = \frac{8}{3-2} + 2 = 10$$

Therefore, $z = 10$.

 yz

____   = 4     →   yz = 4( y + z)  →  yz - 4y = 4z  → y (z-4) = 4z  → y = 4z / (z - 4)      

y + z

  xz

____  = 2  →  xz = 2(x + z) → xz- 2x = 2z  → x ( z - 2) = 2z → x = 2z / (z-2)   

x + z

xy

_______ = 1   →   xy  = x + y       (1)

x + y 

So, using (1)

4z / (z -4)  *  2z / (z -2)  =    4z / (z-4)   +  2z/(z -2)           mulyiply through by (z -4) (z-2)

4z * 2z  =  4z (z - 2) + 2z (z-4)

8z^2   =  4z^2 - 8z + 2z^2 - 8z

2z^2 + 16z  = 0

z^2 +8z  = 0

z ( z + 8)  = 0

z = 0   (reject)

z =  - 8