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# Help on question

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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

### 2+0 Answers

#1
+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
+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

Jun 1, 2024