Introductory Algebra for College Students (7th Edition)

The solution is: $$x = 3$$ To check if this solution is correct, we plug $3$ into the original equation to see if both sides equal one another: $$\frac{(3)(3)}{5} − \frac{2}{5} = \frac{3}{3} + \frac{2}{5}$$ We multiply by the least common denominator of all the fractions: $$15(\frac{9}{5} − \frac{2}{5}) = 15(\frac{3}{3} + \frac{2}{5})$$ $$27 - 6 = 15 + 6$$ Simplifying, we get: $$21 = 21$$ 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 $15$. We now multiply both sides by $15$ to get: $$15(\frac{3x}{5} − \frac{2}{5}) = 15(\frac{x}{3} + \frac{2}{5})$$ We now use the distributive property to simplify the equation: $$15(\frac{3x}{5}) − 15(\frac{2}{5}) = 15(\frac{x}{3}) + 15(\frac{2}{5})$$ Divide out the common factors to get rid of the fractions: $$9x − 6 = 5x + 6$$ Subtract $5x$ from each side to isolate the variable on one side: $$4x − 6 = 6$$ Add $6$ to both sides to separate the variable and the constant terms: $$4x = 12$$ $$x = 3$$ To check if this solution is correct, we plug $3$ into the original equation to see if both sides equal one another: $$\frac{(3)(3)}{5} − \frac{2}{5} = \frac{3}{3} + \frac{2}{5}$$ We multiply by the least common denominator of all the fractions: $$15(\frac{9}{5} − \frac{2}{5}) = 15(\frac{3}{3} + \frac{2}{5})$$ $$27 - 6 = 15 + 6$$ Simplifying, we get: $$21 = 21$$ We see that both sides are equal, so we know that the solution is correct.