The numerator of a fraction is 8 less than the denominator. If the denominator is increased by 5 and the numerator is increased by 3, we get the fraction \(\tfrac34\). Find the original fraction.
\(\tfrac{27}{35}\)
Let the denominator be d, so the numerator is d-8.
New fraction: \(\dfrac{(d-8)+3}{d+5} = \dfrac34 \Rightarrow \dfrac{d-5}{d+5} = \dfrac34\).
\(4(d-5) = 3(d+5) \Rightarrow 4d-20 = 3d+15 \Rightarrow d = 35\).
Numerator: \(35-8 = 27\).
Hence, the original fraction is \(\tfrac{27}{35}\).
Simplify: $3((\frac{5}{3})x^2 - 28x + 15) - 5(x^2 + 6x - 15)$
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