Solving $x,y,z$: Equations (1), (2), (3)

[Albert1]

find solutions of $x,y,z$

$\dfrac {1}{x}+\dfrac {1}{y+z}=\dfrac {1}{2}-----(1)$

$\dfrac {1}{y}+\dfrac {1}{z+x}=\dfrac {1}{3}-----(2)$

$\dfrac {1}{z}+\dfrac {1}{x+y}=\dfrac {1}{4}-----(3)$

 

[soroban]

Hello, Albert!

I have a very looong solution.

Find solutions of $x,y,z$

.$$\begin{array}{cccc}\frac {1}{x}+\frac {1}{y+z} &=& \frac {1}{2} & [1[ \\ \frac {1}{y}+\frac {1}{z+x}&=&\frac {1}{3} & [2] \\ \frac {1}{z}+\frac {1}{x+y}&=&\frac {1}{4} & (3) \end{array}$$

Note that: .$$x,y,z\, \ne\,0.$$

$$\begin{array}{cccccccccc}[1]\!:\;2y + 2x + 2x \:=\:x(y+z) &\Rightarrow& x + y + z \:=\: \frac{xy+xz}{2} & [4] \\ [2]\!:\; 3x + 3x + 3y \:=\:y(x+z) &\Rightarrow & x+y+z \:=\:\frac{xy+yz}{3} & [5] \\ [3]\!:\; 4x+3y+4z \:=\:z(x+y) & \Rightarrow & x+y+z \:=\:\frac{xz+yz}{4} & [6] \end{array}$$

$$\text{From }[4],[5],[6]\!:\;\underbrace{\frac{xy+xz}{2}}_{[7]} \:=\:\underbrace{\frac{xy + yz}{3}}_{[8]} \:=\:\underbrace{\frac{xz+yz}{4}}_{[9]}$$

$$[7]=[8]\!:\;3xy + 3xz \:=\:2xy + 2yz \;\;\;\Rightarrow\;\;\; x \:=\:\frac{2yz}{y+3x}\;\;[10]$$

$$[8] = [9]\!:\;4xy + 4yz \:=\:3xz + 3yz \;\;\;\Rightarrow\;\;\;x \:=\:\frac{yz}{3x-4y}\;\;[11]$$

$$[10]=[11]\!:\;\frac{2yz}{y+3x} \:=\:\frac{yz}{3z-4y} \quad\Rightarrow\quad z \:=\:3y\;\;[12]$$

Substitute into [10]: .$$x \:=\:\frac{2y(3y)}{y+3(3y)} \quad\Rightarrow\quad x \:=\:\tfrac{3}{5}y\;\;[13]$$

Substitute [12] and [13] into [4]:
.. $$2y + 2(3y) + 2\left(\tfrac{3}{5}y\right) \;=\; \left(\tfrac{3}{5}y\right)y + \left(\tfrac{3}{5}y\right)(3y)$$
. . $$\tfrac{46}{5}y \:=\:\tfrac{12}{5}y^2 \quad\Rightarrow\quad 6y^2\:=\:23y \quad\Rightarrow\quad \boxed{y \:=\:\tfrac{23}{6}}$$

Substitute onto [13}: .$$x \:=\:\tfrac{3}{5}\left(\tfrac{23}{6}\right) \quad\Rightarrow\quad \boxed{x \:=\:\frac{23}{10}}$$

Substitute into [12]: .$$z \:=\:3\left(\tfrac{23}{6}\right) \quad\Rightarrow\quad \boxed{z \:=\:\frac{23}{2}}$$

 

[Opalg]

http://www.mymathforum.com/viewtopic.php?t=36640&p=151646

 

[Albert1]

http://www.mymathforum.com/viewtopic.php?t=36640&p=151646

Nicely done!
very good solution there :)

 

[CellCoree]

I would first analyze the equations and their relationships to each other. It is clear that these equations form a system of linear equations, with three unknown variables: x, y, and z. In order to solve for these variables, we would need to use methods such as elimination or substitution.

However, before proceeding with solving the equations, we should check for any inconsistencies or contradictions within the system. This can be done by rearranging the equations and looking for any patterns or relationships between them. For example, in this system, we can see that the denominators of each equation are all related to the variables in the other equations. This suggests that there may be a unique solution to this system of equations.

Next, we can use algebraic methods such as elimination or substitution to solve for the variables. By rearranging the equations and substituting one into the other, we can eventually solve for each variable. It is important to note that there may be multiple solutions to this system of equations, so we should check our answers and make sure they satisfy all three equations.

In conclusion, as a scientist, I would carefully analyze the system of equations and use appropriate methods to solve for the variables. I would also check for any inconsistencies or contradictions and make sure to verify my solutions. By doing so, we can confidently find the solutions for x, y, and z.