Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 2 - Review Exercises - Page 203: 25

Answer

The solution is: $$x = 2$$ To check if the solution is correct, we plus $2$ in for $x$ into the original equation: $$\frac{(2)(2)}{3} = \frac{2}{6} + 1$$ Multiply all terms by their least common denominator, which is $6$ in this case, to get rid of the fractions: $$6(\frac{4}{3}) = 6(\frac{2}{6}) + 6(1)$$ Simplify by dividing out common factors: $$2(4) = 2 + 6$$ $$8 = 8$$ Both sides of the equation are equal, so the solution is correct.

Work Step by Step

We want to multiply all terms by their least common denominator, which is $6$ in this case, so we don't have to deal with fractions. $$6(\frac{2x}{3}) = 6(\frac{x}{6}) + 6(1)$$ We divide out common factors to get rid of the fractions: $$4x = x + 6$$ We subtract $x$ from both sides to isolate the variable: $$3x = 6$$ Divide by $3$ on both sides to isolate $x$: $$x = 2$$ To check if the solution is correct, we plus $2$ in for $x$ into the original equation: $$\frac{(2)(2)}{3} = \frac{2}{6} + 1$$ Multiply all terms by their least common denominator, which is $6$ in this case, to get rid of the fractions: $$6(\frac{4}{3}) = 6(\frac{2}{6}) + 6(1)$$ Simplify by dividing out common factors: $$2(4) = 2 + 6$$ $$8 = 8$$ Both sides of the equation are equal, so the solution is correct.
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