document.write( "Question 1194064: The following augmented matrix is in row echelon form and represents a linear system. Use​ back-substitution to solve the system if possible.
\n" ); document.write( "[1 1 -1|2]
\n" ); document.write( "[0 1 -1|2]
\n" ); document.write( "[0 0 1|1]
\n" ); document.write( "What is the solution to the linear​ system?
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Algebra.Com's Answer #826159 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "The bottom row indicates we have this equation
\n" ); document.write( "0x+0y+1z = 1
\n" ); document.write( "which simplifies to
\n" ); document.write( "z = 1\r
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\n" ); document.write( "\n" ); document.write( "The middle row says
\n" ); document.write( "0x+1y+(-1)z = 2
\n" ); document.write( "which simplifies to
\n" ); document.write( "y-z = 2\r
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\n" ); document.write( "\n" ); document.write( "Plug in z = 1 and solve for y
\n" ); document.write( "y-z = 2
\n" ); document.write( "y-1 = 2
\n" ); document.write( "y = 2+1
\n" ); document.write( "y = 3\r
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\n" ); document.write( "\n" ); document.write( "The first line of the given augmented matrix says we have this equation
\n" ); document.write( "1x+1y+(-1)z = 2
\n" ); document.write( "aka
\n" ); document.write( "x+y-z = 2\r
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\n" ); document.write( "\n" ); document.write( "Plug in y = 3 and z = 1. Solve for x.
\n" ); document.write( "x+y-z = 2
\n" ); document.write( "x+3-1 = 2
\n" ); document.write( "x+2 = 2
\n" ); document.write( "x = 2-2
\n" ); document.write( "x = 0\r
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\n" ); document.write( "\n" ); document.write( "Solution:
\n" ); document.write( "x = 0, y = 3, z = 1
\n" ); document.write( "We can condense this into the ordered triple (x,y,z) = (0,3,1)
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