Introductory Algebra for College Students (7th Edition)

The solution is: $$z = 15$$ To check if this solution is correct, we plug $15$ into the original equation to see if both sides equal one another: $$\frac{15}{5} - \frac{1}{2} = \frac{15}{6}$$ Simplifying, we get: $$3 - \frac{1}{2} = \frac{15}{6}$$ Now we simplify the left side of the equation by changing $3$ into an improper fraction so we can subtract $\frac{1}{2}$ from it: $$\frac{6}{2} - \frac{1}{2} = \frac{15}{6}$$ $$\frac{5}{2} = \frac{15}{6}$$ We can now reduce the fraction on the right-hand side of the equation by dividing both numerator and denominator by $3$ to get: $$\frac{5}{2} = \frac{5}{2}$$ 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 in the equation, which, in this case, is $30$. We now multiply both sides by $30$ to get: $$30(\frac{z}{5} - \frac{1}{2}) = 30(\frac{z}{6})$$ We now use the distributive property to simplify the equation: $$\frac{30z}{5} - \frac{30}{2} = \frac{30z}{6}$$ Divide out the common factors to get rid of the fractions: $$6z - 15 = 5z$$ Subtract $5z$ from each side to isolate the variable on one side: $$z - 15 = 0$$ We can now solve for $z$ by adding $15$ to each side: $$z = 15$$ To check if this solution is correct, we plug $15$ into the original equation to see if both sides equal one another: $$\frac{15}{5} - \frac{1}{2} = \frac{15}{6}$$ Simplifying, we get: $$3 - \frac{1}{2} = \frac{15}{6}$$ Now we simplify the left side of the equation by changing $3$ into an improper fraction so we can subtract $\frac{1}{2}$ from it: $$\frac{6}{2} - \frac{1}{2} = \frac{15}{6}$$ $$\frac{5}{2} = \frac{15}{6}$$ We can now reduce the fraction on the right-hand side of the equation by dividing both numerator and denominator by $3$ to get: $$\frac{5}{2} = \frac{5}{2}$$ We see that both sides are equal, so we know that the solution is correct.