# Solving $x$, $y$ and $z$

If I have $22x = 23y = 24z$ and $x+y+z = 3865$ how to obtain the values of $x$,$y$ and $z$ ?

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How about substitution? –  Andy Jan 2 '11 at 11:25
Indeed substitution works like a charm :) –  Quixotic Jan 2 '11 at 11:57

use $y = \frac{22}{23}x$ and $z = \frac{22}{24}x$ in last equation.
You can use $\TeX$ markup to beautify your equations by putting $-marks around them. – FUZxxl Jan 2 '11 at 11:29 got it, thanks. – sam Jan 2 '11 at 11:44 also you can right-click on any$\TeX$, select Show Source, and see how it was done. FUZxxl got his fractions using the \frac command. For this answer, they are clear enough, but it is a good way to learn. – Ross Millikan Jan 2 '11 at 11:59 cool...................... – sam Jan 2 '11 at 12:28 First, try to express$y$and$z$by$x:$$$22x = 23y\Leftrightarrow y = \frac{22}{23}x$$ $$22x = 24z\Leftrightarrow z = \frac{11}{12}x$$ Then put this into your term to archieve a value for$x:$$$x+y+z=3865\Leftrightarrow x+\frac{22}{23}x+\frac{11}{12}x=3865$$ Now you have the value of$x.$To figure out$y$and$z\$ is left as an exercise to you.