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I only got to getting rid of the denominator and turning the equation into 16x-3 = ax^2+a+bx^2+cx, but from that on I don't know what to do.

Asha R
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    For the equation to be true for all real $x$, except for those which are not applicable, i.e., $x = 0$ due to it being used in $2$ denominators, what must the coefficients of each power of $x$ (including the $0$'th power, i.e., the constant) in your resulting equation be when they are all to one side of the equation with the other side being $0$? – John Omielan May 21 '22 at 21:45
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    @JohnOmielan Thank you! The coefficients must all be 0. – Asha R May 21 '22 at 22:09

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Rearrange $16x-3 = ax^2+a+bx^2+cx$ into a standard-form quadratic equation.

$$(a + b) x^2 + (c - 16)x + (a + 3) = 0$$

Since this must be true for all $x$, all of the coefficients must be zero, i.e., $a + b = 0$, $c - 16 = 0$, and $a + 3 = 0$. The solution is $a = -3, b = 3, c = 16$.

Dan
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