# SOLUTION: I'm just stuck on this easy question. I know I just use the LCD; but I keep getting stuck. (x +1) + (x+2)=2 The (x+1) is over 3 and the (x+2) is over 7, so they are two fractions

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: I'm just stuck on this easy question. I know I just use the LCD; but I keep getting stuck. (x +1) + (x+2)=2 The (x+1) is over 3 and the (x+2) is over 7, so they are two fractions       Log On

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 Click here to see ALL problems on Linear Algebra Question 25850: I'm just stuck on this easy question. I know I just use the LCD; but I keep getting stuck. (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. I have tried multiplying the whole equation by 21; isn' that correct? i need some help.Answer by elima(1433)   (Show Source): You can put this solution on YOUR website!+=2 The LCD=21, so you multiply the first equation by 7, the second equation by 3; +=2 +=2 =2 =2 Now multiply each side by 21; 10x+13=42 10x=42-13 10x=29 x=2.9 Hope this helps =)