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The system of equations \[\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4\] has exactly one solution. What is $z$ in this solution?
 Jun 1, 2024
 #1
avatar+677 
+1

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, \(z = 10\).

 Jun 1, 2024
 #2
avatar+129747 
+1

 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

 

cool cool cool

 Jun 1, 2024

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