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Algebra Difficulty 1.4 Junior Find the answer Canada

If $a=\dfrac{2}{3}b$ and $b \neq 0$ , then $\dfrac{9a+8b}{6a}$ is equal to

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Solution

Since $a = \dfrac{2}{3}b$ , then $3a = 2b$ . Since $b \neq 0$ , then $a \neq 0$ .

Thus, $\dfrac{9a+8b}{6a} = \dfrac{9a+ 4(2b)}{6a} = \dfrac{9a+4(3a)}{6a}=\dfrac{21a}{6a} = \dfrac{7}{2}$ (since $a \neq 0$ ).

Alternatively, $\dfrac{9a+8b}{6a} = \dfrac{3(3a)+ 8b}{2(3a)} = \dfrac{3(2b)+8b}{2(2b)}=\dfrac{14b}{4b} = \dfrac{7}{2}$ (since $b \neq 0$ ).

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.