Introductory Algebra for College Students (7th Edition)

The solution is: $$y = 1$$ To check if this solution is correct, we plug $1$ into the original equation to see if both sides equal one another: $$\frac{1}{12} + \frac{1}{6} = \frac{1}{2} - \frac{1}{4}$$ We again multiply by the least common denominator ($12$) to get rid of the fractions: $$12(\frac{1}{12} + \frac{1}{6}) = 12(\frac{1}{2} - \frac{1}{4})$$ We can now divide out common factors to get rid of the fractions: $$1 + 2 = 6 - 3$$ Simplifying, we get: $$3 = 3$$ We see that both sides are equal, so we know that the solution is correct.
To solve this equation, we want to transform it so that we don't have to work with fractions. We find the least common denominator for all fractions, which, in this case, is $12$. We now multiply both sides by $12$ to get: $$12(\frac{y}{12} + \frac{1}{6}) = 12(\frac{y}{2} - \frac{1}{4})$$ We now use the distributive property to simplify the equation: $$12(\frac{y}{12}) + 12(\frac{1}{6} = 12(\frac{y}{2}) -12(\frac{1}{4}$$ Divide out the common factors to get rid of the fractions: $$y + 2 = 6y - 3$$ Subtract $y$ from each side to isolate the variable on one side: $$2 = 5y-3$$ Add $3$ to both sides to separate the variables and constants: $$5y = 5$$ We can now solve for $y$: $$y = 1$$ To check if this solution is correct, we plug $1$ into the original equation to see if both sides equal one another: $$\frac{1}{12} + \frac{1}{6} = \frac{1}{2} - \frac{1}{4}$$ We again multiply by the least common denominator ($12$) to get rid of the fractions: $$12(\frac{1}{12} + \frac{1}{6}) = 12(\frac{1}{2} - \frac{1}{4})$$ We can now divide out common factors to get rid of the fractions: $$1 + 2 = 6 - 3$$ Simplifying, we get: $$3 = 3$$ We see that both sides are equal, so we know that the solution is correct.