# Multivariable linear equation with 3 variables

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.

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.

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*}

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.

• +1 for the logarithm idea, even though it might be overkill. May 9, 2017 at 16:16
• @AlgorithmsX: Overkill ? In the end the solution is a product of powers (possibly noninteger), which are computed... by logarithms.
– user65203
May 9, 2017 at 17:30
• $\sqrt y=v\implies y=v^2$. 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. May 9, 2017 at 17:35