document.write( "Question 107637: find all the values of x satisfying the given conditions
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document.write( " 1 2 -4
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document.write( "y1= --- , ----, ---------, and y1+y2=y3
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document.write( " x+7 x+3 x^2+10+21\r
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Algebra.Com's Answer #78402 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! Your problem as written doesn't make sense, but I'm going to make an assumption that does and go with that. What I think you meant was:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "That means we can say:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "To add the two terms on the left side, we need a Lowest Common Denominator, given by:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now we have:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Simplify:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since the denominators are equal on both sides, the numerators must be equal as well for the statement to be true, therefore:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But that causes a problem because -7 is not in the domain of either the y1 or y3 functions. x=-7 leads to a zero denominator in each case. Therefore, there are no values of x that satisfy the given conditions. Presuming my assumptions about the construct of the original problem were correct, that is. \n" ); document.write( " |