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How can I solve this system of equations of variables x, y and z: $$ xy -2 \sqrt y + 3yz = 8 \\ 2xy -3 \sqrt y + 2yz = 7\\ -xy + \sqrt y + 2yz = 4 $$

I'm used to solve problems with singular variables ( like 2x +3y-5z= k), and I saw this problem on an exam I want to aply. Thanks for giving me at least one idea how to solve it.

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3 Answers 3

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Let $u=xy$, $v=\sqrt y$, and $w=yz$. Solve for $u,\ v$, and $w$. Then, solve for your original variables, as you'll have: $$\begin{align} xy&=u\\ \sqrt y&=v\\ yz&=w \end{align} $$ and you'll know the values of the three variables on the right. From there, solve for $y$ first. and use substitution to solve the rest.

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Can you solve the following linear system linear system in $a_1,a_2,a_3 \in \mathbb{R}$? \begin{align*} a_1-2a_2+3a_3 &= 8 \\ 2a_1 - 3a_2 + 2a_3 &= 7 \\ -a_1 + a_2 + 2a_3 &= 4 \end{align*}

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The resolution of the "linear" system leads to a solution of the form

$$\begin{cases}xy&=a,\\\sqrt y&=b,\\yz&=c.\end{cases}$$

Now taking the logarithm and using uppercase to denote it, this is equivalent to

$$\begin{cases}X+Y&=A,\\\dfrac12Y&=B,\\Y+Z&=C,\end{cases}$$ another linear system.

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  • $\begingroup$ +1 for the logarithm idea, even though it might be overkill. $\endgroup$
    – AlgorithmsX
    May 9, 2017 at 16:16
  • $\begingroup$ @AlgorithmsX: Overkill ? In the end the solution is a product of powers (possibly noninteger), which are computed... by logarithms. $\endgroup$
    – user65203
    May 9, 2017 at 17:30
  • $\begingroup$ $\sqrt y=v\implies y=v^2$. This is a linear system, so $u,\ v, \text{and }w$ are all rational. This means $y$ is rational because the square of a rational number is rational. Finally, this means $x$ and $z$ are rational because the product of rational numbers is a rational number. You don't need logarithms in this case, but I could see this being necessary in more complicated situations. $\endgroup$
    – AlgorithmsX
    May 9, 2017 at 17:35

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