document.write( "Question 527446: the sum of the the digits of a two-digit number is 10. The tens digit is 2 less than the square of the units digit. Find the number. \n" ); document.write( "
Algebra.Com's Answer #348904 by mathstutor494(120)\"\" \"About 
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Let unit digit be y and tens digit be x
\n" ); document.write( "Therefore the number is 10x+y\r
\n" ); document.write( "\n" ); document.write( "Further, given that sum of the the digits of a two-digit number is 10
\n" ); document.write( "So x+y =10 ...............................................(1)\r
\n" ); document.write( "\n" ); document.write( "Also given that the tens digit is 2 less than the square of the units digit
\n" ); document.write( "So x= y^2-2...............................................(2)\r
\n" ); document.write( "\n" ); document.write( "Substituting value of x from eq 1 in eq 2 results in
\n" ); document.write( "10-y = y^2 -2
\n" ); document.write( "Rearranging above terms of the eq results in
\n" ); document.write( "y^2+y-12=0
\n" ); document.write( "y^2+4y-3y-12=0
\n" ); document.write( "y(y+4)-3(y+4)=0
\n" ); document.write( "(Y-3)(Y+4)=0
\n" ); document.write( "So y = 3 or -4
\n" ); document.write( "As the digit can not be negative, therefore y=3\r
\n" ); document.write( "\n" ); document.write( "Hence x=7 (from eq 1)\r
\n" ); document.write( "\n" ); document.write( "Therefore the number is = 7*10+3
\n" ); document.write( "Ans = 73\r
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