document.write( "Question 935699: A number is 4 more than another. If thrice the larger number is divided by the smaller, the quotient is 5 and the remainder is 6. What are the numbers? \n" ); document.write( "
Algebra.Com's Answer #568996 by Theo(13342)\"\" \"About 
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x is the larger number.
\n" ); document.write( "y is the smaller number.
\n" ); document.write( "x = y + 4 means that the larger number is 4 more than the smaller number.
\n" ); document.write( "3x/y = 5 with a remainder of 6 means three times the larger number divided by the smaller number is equal to 5 with a remainder of 6.
\n" ); document.write( "3x/y = 5 with a remainder of 6 can also be shown as 3x/y = 5 + 6/y.
\n" ); document.write( "you have 2 equations that need to be solved simultaneously.
\n" ); document.write( "they are:
\n" ); document.write( "x = y + 4
\n" ); document.write( "3x/y = 5 + 6/y
\n" ); document.write( "replace x with y + 4 in the second equation to get:
\n" ); document.write( "3(y+4)/y = 5 + 6/y
\n" ); document.write( "multiply both sides of this equation by y to get:
\n" ); document.write( "3(y+4) = 5y + 6
\n" ); document.write( "distribute the multiplication to get:
\n" ); document.write( "3y + 12 = 5y + 6
\n" ); document.write( "subtract 3y from both sides of this equation and subtract 6 from both sides of this equation to get:
\n" ); document.write( "6 = 2y
\n" ); document.write( "solve for y to get:
\n" ); document.write( "y = 3
\n" ); document.write( "if y = 3, then x must be equal to 4.
\n" ); document.write( "that's your solution.
\n" ); document.write( "you need to go back to the original equations to confirm the solution is correct.
\n" ); document.write( "x = y + 4 becomes 7 = 3 + 4 which is correct.
\n" ); document.write( "3x/y = 5 + 6/y becomes 3(7)/3 = 5 + 6/3 which becomes 21/3 = 5 + 2 which becomes 7 = 7 which is also correct.
\n" ); document.write( "the solution looks good.
\n" ); document.write( "there was a wrinkle in there because 21/3 = 7 with a remainder of 0.
\n" ); document.write( "that would cause confusion and make you think that the solution is not correct, but:
\n" ); document.write( "21/3 = 5 with a remainder of 6 is equivalent to 21/3 = 7, so the solution is correct even if the confirmation looks odd.\r
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