Math CentralQuandaries & Queries


Question from jenna, a student:


Hi Jenna,

I assume you are to solve for $y.$

I would eliminate the fractions to start with. To do this multiply both sides b4 $3 \times 2 = 6.$ this gives

\[6 \times \frac{2y + 1}{3} = 6 \times \frac{1y - 1}{2}\]

which simplifies to

\[2 \times (2y + 1) = 3 \times (y - 1)\]

Can you complete it from here? Write back if you need more assistance,

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