document.write( "Question 1124084: The sum of three numbers is 14. The sum of twice the first​ number, 4 times the second​ number, and 5 times the third number is 62. The difference between 4 times the first number and the second number is negative 1. Find the three numbers.\r
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Algebra.Com's Answer #740456 by ikleyn(52776)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let x, y and z be the 1-st, the 2-nd and the 3-rd numbers, respectively.\r\n" );
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document.write( "Then from the condition you have these three equations in 3 unknowns\r\n" );
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document.write( " x +  y +  z = 14,        (1)\r\n" );
document.write( "2x + 4y + 5z = 62,        (2)\r\n" );
document.write( "4x -  y      = -1.        (3)\r\n" );
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document.write( "From eq(3), express  y = 4x +1  and substitute it into eq(1) and eq(2). You will get\r\n" );
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document.write( " x + (4x+1)  +  z = 14    (1')\r\n" );
document.write( "2x + 4(4x+1) + 5z = 62    (2')\r\n" );
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document.write( "Simplify and write it in the standard form\r\n" );
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document.write( " 5x +  z = 13             (1'')\r\n" );
document.write( "18x + 5z = 58             (2'')\r\n" );
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document.write( "Congratulations !  You reduced the 3x3-system to 2x2-system, which is much easier to solve.\r\n" );
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document.write( "To solve it, apply Elimination method.  For it, multiply eq(1'') by 5  (both sides). You will get\r\n" );
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document.write( "25x + 5z = 65             (1''')\r\n" );
document.write( "18x + 5z = 58.            (2''')\r\n" );
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document.write( "Now subtract eq((1''')  from eq(2'''). You will get\r\n" );
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document.write( "7x = 65 - 58 = 7  ====>  x = 1.\r\n" );
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document.write( "Then from eq(1'')  z = 13-5x = 13 - 5*1 = 8.\r\n" );
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document.write( "Finally,  y = 14 - x - z = 14 - 1 - 8 = 5.\r\n" );
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document.write( "Answer.  x = 1;  y = 5;  z = 8.\r\n" );
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