document.write( "Question 1060139: y=3x+2
\n" );
document.write( "3x+y=1 \n" );
document.write( "
Algebra.Com's Answer #675115 by math_helper(2461)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Since you have 2 equations and 2 unknowns, the first thing is to try to solve for one of the variables. So long as the two equations are \"linearly independent\" (or not linearly-dependent) there will be a solution. \n" ); document.write( "Linearly dependent means one equation can be re-arranged to exactly match the other (in essence, if they only \"look\" like two different equations but are in fact the same). Ok, enough on that… \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(1) conveniently expresses y as a function of x, so let's substitute \"3x+2\" from (1) for \"y\" in (2):\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If \n" ); document.write( "\n" ); document.write( "—\r \n" ); document.write( "\n" ); document.write( "Ans: \n" ); document.write( "—\r \n" ); document.write( "\n" ); document.write( "Check: \n" ); document.write( " (1) Does 3/2 = (3*(-1/6) + 2)? \n" ); document.write( " 3/2 = (-3/6)+(12/6) \n" ); document.write( " 3/2 = 9/6 \n" ); document.write( " 3/2 = 3/2 (ok) \n" ); document.write( "- \n" ); document.write( " (2) Does 3(-1/6) + (3/2) = 1 ? \n" ); document.write( " -(1/2) + (3/2) = 1 ? \n" ); document.write( " 2/2 = 1 (ok)\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |