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Solve

xy + x + y = 23

yz + y + z = 31

zx + z + x = 47

in real numbers.

 May 16, 2022
 #1
avatar+9461 
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Add 1 to each equation.

 

\(\begin{cases}xy + x + y + 1 = 24\\yz + y + z + 1= 32\\zx + z + x + 1 = 48\end{cases}\)

 

We do this because now the left-hand side is factorable.

 

\(\begin{cases}(x + 1)(y + 1) = 24\qquad --- (1)\\(y + 1)(z + 1) = 32\qquad --- (2)\\(z + 1)(x + 1) = 48\qquad --- (3)\end{cases}\)

 

Now consider (1) * (2) / 3, we have

 

\(\dfrac{(x + 1)(y + 1)^2(z + 1)}{(z + 1)(x + 1)} = \dfrac{24 \cdot 32}{48}\\ (y + 1)^2 = 16\\ y = 3\text{ or }y = -5\)

 

When y = 3,

(x + 1)(4) = 24

x = 5

(z + 1)(5 + 1) = 48

z = 7

 

Therefore (x, y, z) = (5, 3, 7) is a solution.

 

When y = -5,

(x + 1)(-4) = 24

x = -7

(z + 1)(-7 + 1) = 48

z = -9

 

Therefore (x, y, z) = (-7, -5, -9) is another solution.

 

Solutions are (x, y, z) = (5, 3, 7) and (x, y, z) = (-7, -5, -9).

 May 16, 2022

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