Answer :
count "x" from first equation:
x+4y-5z=-7
x=5z-4y-7
count "x" from second equation:
[tex]3x+2y+3z=7 \\ 3x=7-2y-3z \qquad /:3 \\ x=\frac{7-2y-3z}{3}[/tex]
so:
[tex]5z-4y-7=\frac{7-2y-3z}{3} \qquad /\cdot 3 \\ 15z-12y-21=7-2y-3z \\ 18z-10y=28 \qquad /:2 \\ 9z-5y=14 \\ 9z=14+5y \qquad /:9 \\ z=\frac{14+5y}{9}[/tex]
Now substitute this value of "z" to "x=5z-4y-7":
[tex]x=5 \cdot \frac{14+5y}{9} -4y-7=\frac{70+25y}{9}-\frac{36y}{9}-\frac{63}{9}=\frac{7-11y}{9}[/tex]
Substitute values "x" and "z" to third equation and evaluate "y":
[tex]2x+y+5z=8 \\ 2 \cdot \frac{7-11y}{9} +y+5 \cdot \frac{14+5y}{9}=8 \qquad /\cdot 9 \\ 2(7-11y)+9y+5(14+5y)=72 \\ 14-22y+9y+70+25y=72 \\ 12y+84=72 \\ 12y=-12 \qquad /:12 \\ y=-1 \\ \hbox{So:} \\ x=\frac{7+11}{9}=\frac{18}{9}=2 \\ z=\frac{14-5}{9}=\frac{9}{9}=1[/tex]
Solution is:
[tex]\begin{cases} x=2 \\ y=-1 \\ z=1 \end{cases}[/tex]
x+4y-5z=-7
x=5z-4y-7
count "x" from second equation:
[tex]3x+2y+3z=7 \\ 3x=7-2y-3z \qquad /:3 \\ x=\frac{7-2y-3z}{3}[/tex]
so:
[tex]5z-4y-7=\frac{7-2y-3z}{3} \qquad /\cdot 3 \\ 15z-12y-21=7-2y-3z \\ 18z-10y=28 \qquad /:2 \\ 9z-5y=14 \\ 9z=14+5y \qquad /:9 \\ z=\frac{14+5y}{9}[/tex]
Now substitute this value of "z" to "x=5z-4y-7":
[tex]x=5 \cdot \frac{14+5y}{9} -4y-7=\frac{70+25y}{9}-\frac{36y}{9}-\frac{63}{9}=\frac{7-11y}{9}[/tex]
Substitute values "x" and "z" to third equation and evaluate "y":
[tex]2x+y+5z=8 \\ 2 \cdot \frac{7-11y}{9} +y+5 \cdot \frac{14+5y}{9}=8 \qquad /\cdot 9 \\ 2(7-11y)+9y+5(14+5y)=72 \\ 14-22y+9y+70+25y=72 \\ 12y+84=72 \\ 12y=-12 \qquad /:12 \\ y=-1 \\ \hbox{So:} \\ x=\frac{7+11}{9}=\frac{18}{9}=2 \\ z=\frac{14-5}{9}=\frac{9}{9}=1[/tex]
Solution is:
[tex]\begin{cases} x=2 \\ y=-1 \\ z=1 \end{cases}[/tex]
