# Solving and Checking: Adding Fractions

How would one go about solving this type of problem?

The problem is: $$\frac{x}{2} + \frac{8}{3} = \frac{1}{6}\ .$$ How to solve and check?

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One way to do it would be first to eliminate the fractions. A common denominator for everything on both sides is $6$: if you multiply both sides by $6$ you get $$6\Bigl(\frac{x}{2}+\frac{8}{3}\Bigr)=6\Bigl(\frac{1}{6}\Bigr)\ .$$ Multiplying out the brackets, $$6\Bigl(\frac{x}{2}\Bigr)+6\Bigl(\frac{8}{3}\Bigr)=6\Bigl(\frac{1}{6}\Bigr)\ .$$ Simplifying, $$3x+16=1\ ,$$ and hopefully you can take it from here.

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There's two ways of solving this problem.

• Method $1$: You should notice that the LCM (Lowest Common Multiple) of $2$ and $3$ is $6$ which is the denominator on RHS (right-hand side) of the equation. So adding the LHS together you'll get $$\frac{3x+16}{6}=\frac{1}{6}$$ And since the denominators on both side are equal then you can equate the numerator. So you'll end up with $$3x+16=1$$ And then you simply solve for $x$.
• Method $2$: The other way as David answered is to Multiply both sides by $6$ to eliminate the denominator.
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Since the goal is to get something like $x=something$, with $x$ by itself, the first thing I'd do is move the 8/3 to the other side $$\frac{x}{2}=\frac{1}{6}-\frac{8}{3}$$ then you want to get rid of the division by $2$, $$x=2\left(\frac{1}{6}-\frac{8}{3}\right)=\frac{1}{3}-\frac{16}{3}=-\frac{15}{3}=-5.$$

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