We number the formulas as (i), (ii), (iii):

$$ start situations} message {(i)} ~ ~ x+ y-2z= 1 \ message {(ii)} ~ ~ 2x-4y+ z= 0 \ message {(iii)} ~ ~ 2y-3z= -1 end situations} $$

From the 3rd (iii), we separate z:

$ latex z= frac {2} {3} y+ frac {1} {3} $

Substitute z in (ii) to get (ii’):

$$ 2x-4y+ frac {2} {3} y+ frac {1} {3} = 2x-frac {10} {3} y+ frac {1} {3} =0;; message {(ii’)} $$

We likewise alternative z in (i) to get (i’):

$$ x+ y-2( frac {2} {3} y+ frac {1} {3} )=, x-frac {1} {3} y-frac {2} {3} =1;; message {(i’)} $$

From the above formula (i’) we separate y: $latex y= 3x-5 $.

Replacing y in (ii’), we obtain: $latex 17-8x= 0$. Deducting, we are entrusted:

$$ x= frac {17} {8} $$

Substitute the discovered worth of x in (i’) as well as isolate y: $latex frac {17} {8} -frac {1} {3} y-frac {2} {3} =1$

to get:

$$ y= frac {11} {8} $$

Finally, we replace x as well as y in the formula (i) as well as address for z:

$$ frac {17} {8} +frac {11} {8} -2 z= 1$$

$$ z= frac {5} {4} $$

Summarizing, the remedy to the system is:

$$ left[ x=frac{17}{8},y=frac{11}{8},z=frac{5}{4}right] $$



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