document.write( "Question 1094214: Consider the system of linear equations in the variables x,y,z given by:
\n" ); document.write( "ax+y+z=4, x+by+z=3, x+2by+z=4\r
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\n" ); document.write( "No solutions
\n" ); document.write( "Exactly one solution
\n" ); document.write( "Infinite solutions \n" ); document.write( "
Algebra.Com's Answer #708831 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "(1)   ax +   y + z = 4\r\n" );
document.write( "(2)    x +  by + z = 3\r\n" );
document.write( "(3)    x + 2by + z = 4\r\n" );
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document.write( "Subtract (1)-(2)\r\n" );
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document.write( "(4)  (a-1)x + (1-b)y = 1\r\n" );
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document.write( "Subtract (1)-(3)\r\n" );
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document.write( "(5)  (a-1)x + (1-2b)y = 0\r\n" );
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document.write( "Subtract (4)-(5)\r\n" );
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document.write( "    (1-b)y-(1-2b)y = 1\r\n" );
document.write( "        y-by-y+2by = 1\r\n" );
document.write( "(6)             by = 1\r\n" );
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document.write( "Case 1  b=0\r\n" );
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document.write( "Substitute in (6)\r\n" );
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document.write( "         0y = 1 there is no solution.\r\n" );
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document.write( "Assume b≠0, then by (6)\r\n" );
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document.write( "                       y = 1/b\r\n" );
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document.write( "Substitute in (4)\r\n" );
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document.write( "(4)      (a-1)x + (1-b)y = 1\r\n" );
document.write( "     (a-1)x + (1-b)(1/b) = 1\r\n" );
document.write( "        (a-1)x + 1/b - 1 = 1\r\n" );
document.write( "            (a-1)x + 1/b = 2\r\n" );
document.write( "             b(a-1)x + 1 = 2b\r\n" );
document.write( "(7)              b(a-1)x = 2b-1\r\n" );
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document.write( "Case 2   a=1, b≠0\r\n" );
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document.write( "         Then from (7), the left side is 0, so\r\n" );
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document.write( "              2b-1 = 0\r\n" );
document.write( "                2b = 1\r\n" );
document.write( "                 b = 1/2\r\n" );
document.write( "                 y = 1/(1/2) = 2 \r\n" );
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document.write( "Substitute in (4)\r\n" );
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document.write( "(4)      (a-1)x + (1-b)y = 1\r\n" );
document.write( "     (a-1)x + (1-1/2)(2) = 1 \r\n" );
document.write( "       (a-1)x + (1/2)(2) = 1\r\n" );
document.write( "              (a-1)x + 1 = 1\r\n" );
document.write( "                  (a-1)x = 0\r\n" );
document.write( "                      0x = 0\r\n" );
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document.write( "Any value of x will make that true, so there are \r\n" );
document.write( "infinitely many solutions when  a=1, b≠0\r\n" );
document.write( "You can solve for z if you like as it will be defined,\r\n" );
document.write( "and unique, since the denominator b is not 0.\r\n" );
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document.write( "Case 3   a≠1   b≠0\r\n" );
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document.write( "(6)               by = 1\r\n" );
document.write( "                   y = 1/b\r\n" );
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document.write( "(7)          b(a-1)x = 2b-1\r\n" );
document.write( "                   x = (2b-1)/[b(a-1)]\r\n" );
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document.write( "(2)      x +  by + z = 3\r\n" );
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document.write( "You can solve for z if you like as it will be defined,\r\n" );
document.write( "and unique, since no denominator is 0.\r\n" );
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document.write( "So \r\n" );
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document.write( "There are no solutions when b=0\r\n" );
document.write( "There are infinitely many solutions when a=1, b≠0\r\n" );
document.write( "There is one solution if a≠1, b≠0 \r\n" );
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document.write( "Edwin
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