Solve an equation system with three unknown variables

**Anna55** · February 27th 2011, 04:08 AM

Please solve this equation system:
xy+x+y=11
xz+x+z=14
yz+y+z=19

First I factorized everything:
(x+1)(y+1)=12
(x+1)(z+1)=15
(y+1)(z+1)=20

After that I tried the substitution method, however I fail to solve it.
I should I countinue?
Thank you in advance!

**Prove It** · February 27th 2011, 04:17 AM

Divide the second equation by the first to give $\displaystyle \frac{z+1}{y+1} = \frac{5}{4} \implies z + 1 = \frac{5}{4}(y + 1)$ .

Now substitute this into the third equation and you can solve for $\displaystyle y$ , and then use that to solve for $\displaystyle x,z$ .

**Anna55** · February 27th 2011, 04:54 AM

Thank you for your help!
I substituted it into the third equation. My answer was that y can be 1 or -3, however when I looked in the key, y should be 3 or 5. Can you please show be how to do it?

**Prove It** · February 27th 2011, 05:29 AM

I think you'll find that $\displaystyle y = 3$ or $\displaystyle y = -5$ ...

**Soroban** · February 27th 2011, 07:16 AM

Hello, Anna55!

$\text{Solve this system: }\;\begin{array}{cccc}xy+x+y&=&11 & [1] \\ xz+x+z&=&14 & [2] \\ yz+y+z&=&19 & [3] \end{array}$

I tried all kinds of "clever" tricks . . . and failed abysmally.
. . I was reduced to using simple Substitution . . .

$\text{In [3], solve for }z\!:\;\;z \;=\;\dfrac{19-y}{1+y}\;\;[4]$

Substitute into [2]: . $x\left(\dfrac{19-y}{1+y}\right) + x + \dfrac{19-y}{1+y} \:=\:14$

. . $\text{Solve for }y\!:\;\;y \;=\;\dfrac{4x+1}{3}\;\;[5]$

$\text{Substutite into [1]: }\;x\left(\dfrac{4x+1}{3}\right) + x + \dfrac{4x+1}{3} \;=\;11$

This simplifies to: . $x^2 + 2x - 8 \:=\:0 \quad\Rightarrow\quad (x-2)(x+4) \:=\:0$

. . Hence: . $x \;=\;2,\:\text{-}4$

$\text{Substitute into [5]: }\;y \:=\:\begin{Bmatrix}\dfrac{4(2)+1}{3} \:=\:3 \\ \\[-3mm] \dfrac{4(\text{-}4)+1}{3} \;=\;\text{-}5 \end{Bmatrix}$

$\text{Substitute into [4]: }\;z \;=\;\begin{Bmatrix}\dfrac{19-3}{1+3} &=& 4 \\ \\[-3mm] \dfrac{19-(\text{-}5)}{1 - 5} &=& \text{-}6 \end{Bmatrix}$

$\text{Therefore: }\;(x,y,z) \;=\;(2,3,4),\;(\text{-}4,\text{-}5,\text{-}6)$

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