document.write( "Question 1040622: I need help with this problem please:\r
\n" ); document.write( "\n" ); document.write( "x-8/x-5 + 30/x^2-25 = 0\r
\n" ); document.write( "\n" ); document.write( "I'm not sure where to begin on this and any help would be greatly appreciated! \n" ); document.write( "
Algebra.Com's Answer #655463 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
(x-8)/(x-5)+30/(x^2-25)=0
\n" ); document.write( "x^2-25=(x+5)(x-5)
\n" ); document.write( "common denominator for the left side is (x+5)(x-5).
\n" ); document.write( "Multiply the first term by (x+5); so (x-8)(x+5)=x^2-3x-40
\n" ); document.write( "The second term is unchanged, and their sum is x^-3x-40+30=x^2-3x-10.
\n" ); document.write( "That is all over (x+5)(x-5)
\n" ); document.write( "x^2-3x-10=(x-5)(x+2)
\n" ); document.write( "Therefore (x+5)(x+2)/(x+5)(x-5)= (x+2)/x-5), since (x+5) cancels.
\n" ); document.write( "(x+2)/(x-5)=0
\n" ); document.write( "That occurs when x=-2. You can multiply both sides of the equation by the denominator (x-5). That gets rid of the denominator on the left, and 0*(x-5)=0
\n" ); document.write( "-----------------------
\n" ); document.write( "Check
\n" ); document.write( "(-10/-7)+(30/-21)=0
\n" ); document.write( "(10/7)-(10/7)=0, after reducing the second term.
\n" ); document.write( " \n" ); document.write( "
\n" );