I am unable to solve this equation: $$\displaystyle\frac{(12x^2+5x-3)}{(4x+3)} = \frac{(6x^2+13x-5)}{(2x+5)}$$ On solving the LHS and RHS become equal and cancel each other. Much grateful if anyone can help me out. Shas
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First compute the roots of numerators and you will see what is happening. – Claude Leibovici Nov 29 '19 at 10:17
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The equation is not defined for $x=-\frac{3}{4}$ and $x=-\frac{5}{2}$. The equation is fulfilled for all other real $x$.
Dr. Sonnhard Graubner
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The answer given is x = 1. This is a school grade equation. Is there any process to handle this type of equation? – shaswata pal Nov 29 '19 at 10:07
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Yeah, it is $$\frac{(4x+3)(3x-1)}{4x+3}=\frac{(2x+5)(3x-1)}{2x+5}$$ so your answer is not true. – Dr. Sonnhard Graubner Nov 29 '19 at 10:25