Simplify a2−3a+29a2−1  ÷  2a−26a−2\displaystyle \frac{a^2-3a+2}{9a^2-1}\;\div\;\frac{2a-2}{6a-2} Answer: ...

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Worked solution

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1. Factorise all possible numerator and denominator expressions.

a2−3a+29a2−1÷2a−26a−2=(a−1)(a−2)(3a−1)(3a+1)÷2(a−1)2(3a−1)\frac{a^2-3a+2}{9a^2-1}\div\frac{2a-2}{6a-2} =\frac{(a-1)(a-2)}{(3a-1)(3a+1)}\div\frac{2(a-1)}{2(3a-1)}

2. Multiply by the reciprocal and cancel common factors.

(a−1)(a−2)(3a−1)(3a+1)×2(3a−1)2(a−1)=a−23a+1\frac{(a-1)(a-2)}{(3a-1)(3a+1)}\times\frac{2(3a-1)}{2(a-1)} =\frac{a-2}{3a+1}
Answer

13a+1\displaystyle \frac{1}{3a+1}

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