Respuesta :
Given,
The linear equations are,
[tex]\begin{gathered} -a-3b+4c=3 \\ 5a-8b+5c=27 \\ 5a-25+6c=1 \end{gathered}[/tex]Taking equation first as,
[tex]\begin{gathered} -a-3b+4c=3 \\ -a=3-4c+3b \\ a=4c-3b-3\ldots\ldots\ldots\ldots\ldots\text{.(i)} \end{gathered}[/tex]Substituting the value of a from equation (i) to equation second,
[tex]\begin{gathered} 5a-8b+5c=27 \\ 5(4c-3b-3)-8b+5c=27 \\ 20c-15b-15-8b+5c=27 \\ 25c-23b=42 \\ -23b=42-25c \\ b=\frac{25c-42}{23}\ldots\ldots\ldots\ldots\ldots\text{.}(ii) \end{gathered}[/tex]Subsituting the value of b in equation (i),
[tex]\begin{gathered} a=4c-3b-3 \\ a=4c-3(\frac{25c-42}{23})-3 \\ a=\frac{92c-75c+126-69}{23} \\ a=\frac{17c+57}{23}\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots\text{.(i}ii) \end{gathered}[/tex]Subsituting the value of a from equation (iii) to last equation,
[tex]\begin{gathered} 5a-25+6c=1 \\ 5(\frac{17c+57}{23})+6c=1+25 \\ 85c+285+138c=598 \\ 223c=313 \\ c\approx1.40 \end{gathered}[/tex]Suubstituting the value of c in equation (ii),
[tex]\begin{gathered} b=\frac{25(3.42)-12}{23} \\ b\approx3.20 \end{gathered}[/tex]Subsituting the value of c in equation (iii),
[tex]\begin{gathered} a=\frac{17(3.42)-33}{23} \\ a\approx1.09 \end{gathered}[/tex]