document.write( "Question 26100: I need help with the following: (my son asked me to help but I'm VERY rusty)\r
\n" );
document.write( "\n" );
document.write( "3A + B + 4C = 34
\n" );
document.write( "2A + 2B + C = 26
\n" );
document.write( "A + 2B + 3C = 23 \n" );
document.write( "
Algebra.Com's Answer #14030 by askmemath(368)![]() ![]() ![]() You can put this solution on YOUR website! 3 equations and 3 unknowns. Take 2 equations at a time and try to eliminate 1 common variable\r \n" ); document.write( "\n" ); document.write( "3A + B + 4C = 34 (1) \n" ); document.write( "2A + 2B + C = 26 (2) \n" ); document.write( "A + 2B + 3C = 23 (3)\r \n" ); document.write( "\n" ); document.write( "Subtracting (3) fm (2) \n" ); document.write( "A - 2C = 3 (4)\r \n" ); document.write( "\n" ); document.write( "Multiplying (1) with 2 \n" ); document.write( "6A+2B+8C = 68 (5) \n" ); document.write( "Subtracting (2) fm (5) \n" ); document.write( "4A+7C = 42 (6)\r \n" ); document.write( "\n" ); document.write( "Multiplying (4) with 4 \n" ); document.write( "4A-8C = 12 (7)\r \n" ); document.write( "\n" ); document.write( "Subtracting (7) fm (6) \n" ); document.write( "15C = 30 \n" ); document.write( "Dividing by 15 throughout \n" ); document.write( "C = 2\r \n" ); document.write( "\n" ); document.write( "Putting this back in (4) \n" ); document.write( "A - 2(2) = 3 \n" ); document.write( "Adding 4 on both sides \n" ); document.write( "A = 7 \n" ); document.write( "Putting this back in any of (1),(3),(2) \n" ); document.write( "B = 5 \n" ); document.write( " |