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# Solve $\dfrac{10{{x}^{2}}+15x+63}{5{{x}^{2}}-25x+12}=\dfrac{2x+3}{x-5}$

Last updated date: 24th Mar 2023
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Hint: Rather than thinking about dividing we should first consider cross multiplying and see if anything useful comes along. After cross-multiplication, solve using known techniques of algebra to solve for the value of x.

We have the equation- $\dfrac{10{{x}^{2}}+15x+63}{5{{x}^{2}}-25x+12}=\dfrac{2x+3}{x-5}$
After cross-multiplying we get,
$(x-5)(10{{x}^{2}}+15x+63)=(2x+3)(5{{x}^{2}}-25x+12)$
After multiplying the terms in LHS and RHS and writing them separately we get,
$10{{x}^{3}}+15{{x}^{2}}+63x-50{{x}^{2}}-75x-315=10{{x}^{3}}-50{{x}^{2}}+24x+15{{x}^{2}}-75x+36$
We see that the terms present in both LHS and RHS are- $10{{x}^{3}}$ , $-50{{x}^{2}}$ , $-75x$ , $15{{x}^{2}}$
Therefore the following terms cancels out and we are left with
$63x-315=24x+36$
Subtracting 24x both sides we have
$39x-315=36$
Adding 315 both sides we have,
\begin{align} & 39x=351 \\ & \Rightarrow x=\dfrac{351}{39} \\ & \Rightarrow x=9 \\ \end{align}
Therefore, the above question has only one solution that is x=9.
Note: If we would have tried to solve the question any other way it would have been unnecessarily long. If we would have tried to first divide the terms on LHS and RHS and then proceeded it would also have been rather difficult. Someone may also try to factorise the quadratic equation on the LHS first. As we know factorisation also takes time if we cannot easily see how to split the coefficient of x so that it fits for factorisation. Even after we factorise we again would have to multiply if nothing cancels out. We also saved time where we cancelled out $10{{x}^{3}}$ , $-50{{x}^{2}}$ , $-75x$ , $15{{x}^{2}}$ .