From the first and the third equations it follows that z=a−c. Substituting to the second and the third equations we have
x+3y=b−8a+8c and x+2y=c−2a+2c=3c−2a.
Hence y=b−8a+8c−3c+2a=b−6a+5c and x=3c−2a−2b+12a−10c=10a−2b−7c
The answer is
x=10a−2b−7cy=−6a+b+5cz=a−c
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