document.write( "Question 329664: Not sure if I'm in the right place. I am stuck on this problem, and can't find anyone who understands it.\r
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document.write( "3/y-3 - 7/y+2 + 4 = 0\r
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document.write( "If this isn't the right place, can you tell me where I need to post? \r
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document.write( "Thanks so much for your help!!!\r
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document.write( "S\r
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Algebra.Com's Answer #236328 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! solve \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "It's an equation so we can get rid of the denominators by mult by (y-3)(y+2) \n" ); document.write( "(y-3)(y+2)* \n" ); document.write( ": \n" ); document.write( "Cancel the denominators, FOIL \n" ); document.write( "3(y+2) - 7(y-3) + 4(y^2 - y - 6) = 0 \n" ); document.write( ": \n" ); document.write( "3y + 6 - 7y + 21 + 4y^2 - 4y - 24 = 0 \n" ); document.write( ": \n" ); document.write( "Combine like terms, form a quadratic equation \n" ); document.write( "4y^2 + 3y - 7y - 4y + 6 + 21 - 24 = 0 \n" ); document.write( ": \n" ); document.write( "4y^2 - 8y + 3 = 0 \n" ); document.write( "Factor \n" ); document.write( "(2y - 3)(2y - 1) = 0 \n" ); document.write( "Two solutions \n" ); document.write( "2y = 3 \n" ); document.write( "y = 1.5 \n" ); document.write( "and \n" ); document.write( "2y = 1 \n" ); document.write( "y = .5 \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution using x=1.5 in the original equation \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "-2 - 2 + 4 = 0 \n" ); document.write( ": \n" ); document.write( "You can check the x=.5 solution \n" ); document.write( " \n" ); document.write( " |