document.write( "Question 157400: Find two numbers whose sum is 147 and one is 4 greather than three eigths of the other. \n" ); document.write( "
Algebra.Com's Answer #116007 by gonzo(654)\"\" \"About 
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let x = one of the numbers.
\n" ); document.write( "let y be the other number.
\n" ); document.write( "equation for the sum of the numbers being equal to 147 is
\n" ); document.write( "x+y=147
\n" ); document.write( "one of the number is 4 greater than 3/8 times the other number.
\n" ); document.write( "in equation form, this becomes
\n" ); document.write( "\"x+=+%28%283%2F8%29%2Ay%29%2B4\"
\n" ); document.write( "substituting for x in the original equation to eliminate one of the unknowns we get
\n" ); document.write( "x+y=147
\n" ); document.write( "becomes (3/8*y+4)+y=147
\n" ); document.write( "multiplying both sides of the equation by 8 to get rid of the fraction and the equation becomes
\n" ); document.write( "(8*3/8*y) + (8*4) + (8*y) = 8*147
\n" ); document.write( "becomes
\n" ); document.write( "(3*y) + 32 + (8*y) = 1176
\n" ); document.write( "becomes
\n" ); document.write( "11*y = 1144
\n" ); document.write( "becomes
\n" ); document.write( "y = 104
\n" ); document.write( "going back to the original equation of x+y = 147, we get x + 104 = 147 which becomes x = 147 - 104 which becomes x = 43.
\n" ); document.write( "answer is x = 43 and y = 104.
\n" ); document.write( "substituting in the original equation of x+y=147, we get 104+43=147 which becomes 147=147 which confirms the values for x and y are correct for the original equation.
\n" ); document.write( "substituting in the x = 3/8*y + 4 equation, we get 43 = ((3*104)/8)+4 which becomes (312/8) + 4 which becomes 39 + 4 = 43.
\n" ); document.write( "43 = 43 confirms x and y have the correct values.
\n" ); document.write( "answer is:
\n" ); document.write( "x = 43
\n" ); document.write( "y = 104\r
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