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valkas [14]
3 years ago
15

What is the solution of the system of the linear equations?

Mathematics
1 answer:
____ [38]3 years ago
5 0
2x+3y-z=-16 \\
-3x+3y+z=20 \\
-x+2y+z=2 \\ \\
\hbox{the 1st and 2nd equation:} \\
2x+3y-z=-16 \\
\underline{-3x+3y+z=20} \\
2x-3x+3y+3y=20-16 \\
-x+6y=4 \\ \\
\hbox{the 1st and 3rd equation:} \\
2x+3y-z=-16 \\
\underline{-x+2y+z=2 \ \ \ } \\
2x-x+3y+2y=2-16 \\
x+5y=-14

\hbox{the two new equations:} \\
-x+6y=4 \\
\underline{x+5y=-14} \\
6y+5y=4-14 \\
11y=-10 \\
y=-\frac{10}{11} \\ \\
x+5y=-14 \\
x+5 \times (-\frac{10}{11})=-14 \\
x-\frac{50}{11}=-\frac{154}{11} \\
x=-\frac{104}{11} \\
x=-9 \frac{5}{11}

-x+2y+z=2 \\
-(-\frac{104}{11})+2 \times (-\frac{10}{11})+z=2 \\
\frac{104}{11}-\frac{20}{11}+z=\frac{22}{11} \\
\frac{84}{11}+z=\frac{22}{11} \\
z=-\frac{62}{11} \\
z=-5 \frac{7}{11} \\ \\
\boxed{x=-9 \frac{5}{11}, y=-\frac{10}{11}, z=-5 \frac{7}{11}}
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<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>!</em>

4 0
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