document.write( "Question 1090279: x-z=-3
\n" ); document.write( "y+z=9
\n" ); document.write( "x+z=7
\n" ); document.write( "solve the following linear system by using gaussian reduction method matrix \n" ); document.write( "
Algebra.Com's Answer #704901 by greenestamps(13200)\"\" \"About 
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The matrix for the given equations is

\n" ); document.write( "\"matrix%283%2C4%2C1%2C0%2C-1%2C-3%2C0%2C1%2C1%2C9%2C1%2C0%2C1%2C7%29\"

\n" ); document.write( "We first want to get \"1 0 0\" in column 1. It is already nearly the way we want it; we just need to get a 0 in row 3. To do that, we can replace row 3 with (row 3 - row 1).
\n" ); document.write( "R3 <-- R3-R1: 1-1=0; 0-0=0; 1-(-1)=2; 7-(-3)=10:

\n" ); document.write( "\"matrix%283%2C4%2C1%2C0%2C-1%2C-3%2C0%2C1%2C1%2C9%2C0%2C0%2C2%2C10%29\"

\n" ); document.write( "Next we can make smaller numbers in the matrix by dividing row 3 by 2:

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\n" ); document.write( "After this, the second column is already the way we want it. So now we use row 3 to get \"0 0 1\" in column 3.
\n" ); document.write( "R1 <-- R1+R3: 1+0=1; 0+0=0; 1+(-1)=0; -3+5=2
\n" ); document.write( "R2 <-- R2-R3: 0-0=0; 1-0=1; 1-1=0; 9-5=4

\n" ); document.write( "\"matrix%283%2C4%2C1%2C0%2C0%2C2%2C0%2C1%2C0%2C4%2C0%2C0%2C1%2C5%29\"


\n" ); document.write( "The solution to the system of equations is x=2, y=4, z=5.
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