Question

X+4y-5z=-7,3x+2y+3z=7,2x+y+5z=8

Answer 1

count "x" from first equation:
x+4y-5z=-7
x=5z-4y-7
count "x" from second equation:

`3x+2y+3z=7 \\ 3x=7-2y-3z \qquad /:3 \\ x=(7-2y-3z)/(3)`
so:

`5z-4y-7=(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=(14+5y)/(9)`
Now substitute this value of "z" to "x=5z-4y-7":

`x=5 \cdot (14+5y)/(9) -4y-7=(70+25y)/(9)-(36y)/(9)-(63)/(9)=(7-11y)/(9)`
Substitute values "x" and "z" to third equation and evaluate "y":

`2x+y+5z=8 \\ 2 \cdot (7-11y)/(9) +y+5 \cdot (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=(7+11)/(9)=(18)/(9)=2 \\ z=(14-5)/(9)=(9)/(9)=1`
Solution is:

`\begin{cases} x=2 \\ y=-1 \\ z=1 \end{cases}`