document.write( "Question 66341This question is from textbook
\n" ); document.write( ": This is also supposed to be solved using the substitution method.\r
\n" ); document.write( "\n" ); document.write( "r/2 + s/3 = 1
\n" ); document.write( "r/4 + 2s/3 = -1 \n" ); document.write( "
Algebra.Com's Answer #47054 by ankor@dixie-net.com(22740)\"\" \"About 
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r/2 + s/3 = 1
\n" ); document.write( "r/4 + 2s/3 = -1
\n" ); document.write( ":
\n" ); document.write( "First, get rid of those denominators, mult the 1st eq by 6, the 2nd eq by 12
\n" ); document.write( "3r + 2s = 6
\n" ); document.write( "3r + 8s = -12
\n" ); document.write( ":
\n" ); document.write( "Use the 1st eq for substitution for s
\n" ); document.write( "3r + 2s = 6
\n" ); document.write( "2s = 6 - 3r
\n" ); document.write( "divide by 2
\n" ); document.write( "s = 3 - 1.5r
\n" ); document.write( ":
\n" ); document.write( "Substitute (3-1.5r) of s in the 2nd equation:
\n" ); document.write( "3r + 8s = -12
\n" ); document.write( "3r + 8(3-1.5r) = -12
\n" ); document.write( "3r + 24 - 12r = - 12
\n" ); document.write( "3r - 12r = -12 - 24
\n" ); document.write( "Get rid of all those negatives, mult eq by -1
\n" ); document.write( "-3r + 12r = 12 + 24
\n" ); document.write( "9r = 36
\n" ); document.write( "r = 36/9
\n" ); document.write( "r = 4
\n" ); document.write( ":
\n" ); document.write( "Use 3r + 2s = 6 to find s
\n" ); document.write( "3(4) + 2s = 6
\n" ); document.write( "12 + 2s = 6
\n" ); document.write( "2s = 6 - 12
\n" ); document.write( "2s = -6
\n" ); document.write( "s = -3
\n" ); document.write( ":
\n" ); document.write( "Check solutions in the original eq:
\n" ); document.write( "r/2 + s/3 = 1
\n" ); document.write( "4/2 + (-3/3) = 1
\n" ); document.write( "2 - 1 = 1
\n" ); document.write( ":
\n" ); document.write( "You can check our solutions in the original 2nd equation, to be sure\r
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