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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] \$\$