document.write( "Question 95435: =
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document.write( "What is B?\r
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document.write( "I tried the following approach:\r
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document.write( "7=A*(x+9)+B*
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document.write( "7=x*(A+Bx)+9A+5B
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document.write( "7=9A+5B & x(A+Bx)=0\r
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document.write( "I could only say B=(7-9A)/5 & either x=0 or x=-A/B
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document.write( "I'm not sure how to move from here to the next in order to find the solution.\r
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document.write( "Could you help? The question was given by teacher on the board, not from a textbook. \n" );
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Algebra.Com's Answer #69463 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Since the denominator of the 1st fraction is x^2+5 \n" ); document.write( "you need Ax+B as its numerator and C as the numerator of the 2nd fraction \n" ); document.write( "------------------ \n" ); document.write( "Then multiplying thru by (x^2+5)(x+9) you would get: \n" ); document.write( "7 = (Ax+B)(x+9) + C(x^2+5) \n" ); document.write( "7 = Ax^2+9Ax+Bx+9B+Cx^2+5C \n" ); document.write( "7 = (A+C)x^2+(9A+B)x+(9B+5C) \n" ); document.write( "--------------- \n" ); document.write( "Let x=0, then 0A+9B+5C = 7 \n" ); document.write( "Let x=1, then 10A+10B+6C=7 \n" ); document.write( "Let x=-1, then -8A+8B+6C=7 \n" ); document.write( "------------------ \n" ); document.write( "Solving that with a TI-83 Matrix facility you get \n" ); document.write( "A = -0.8139... \n" ); document.write( "B = 0.73255... \n" ); document.write( "C = 0.08139... \n" ); document.write( "============= \n" ); document.write( "That may or may not be what you are looking for but \n" ); document.write( "you cannot arrive at an answer with A as the numerator over \n" ); document.write( "x^2+5 and B as the numerator over x+9 \n" ); document.write( "================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |