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 ->  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


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B1%29%2F3+%28x%2B2%29%2F7=2
The LCD=21, so you multiply the first equation by 7, the second equation by 3;
7%28x%2B1%29%2F21+3%28x%2B2%29%2F21=2
%287x%2B7%29%2F21+%283x%2B6%29%2F21=2
%287x%2B7%2B3x%2B6%29%2F21=2
%2810x%2B13%29%2F21=2
Now multiply each side by 21;
10x+13=42
10x=42-13
10x=29
x=2.9
Hope this helps
=)