document.write( "Question 25850: I'm just stuck on this easy question. I know I just use the LCD; but I keep getting stuck.
\n" ); document.write( "(x +1) + (x+2)=2 The (x+1) is over 3 and the (x+2) is over 7, so they are two fractions being added but I couldn't get it to look right on this screen.\r
\n" ); document.write( "\n" ); document.write( "I have tried multiplying the whole equation by 21; isn' that correct? i need some help.
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Algebra.Com's Answer #13913 by elima(1433)\"\" \"About 
You can put this solution on YOUR website!
\"%28x%2B1%29%2F3\"+\"%28x%2B2%29%2F7\"=2
\n" ); document.write( "The LCD=21, so you multiply the first equation by 7, the second equation by 3;
\n" ); document.write( "\"7%28x%2B1%29%2F21\"+\"3%28x%2B2%29%2F21\"=2
\n" ); document.write( "\"%287x%2B7%29%2F21\"+\"%283x%2B6%29%2F21\"=2
\n" ); document.write( "\"%287x%2B7%2B3x%2B6%29%2F21\"=2
\n" ); document.write( "\"%2810x%2B13%29%2F21\"=2
\n" ); document.write( "Now multiply each side by 21;
\n" ); document.write( "10x+13=42
\n" ); document.write( "10x=42-13
\n" ); document.write( "10x=29
\n" ); document.write( "x=2.9
\n" ); document.write( "Hope this helps
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