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5 ( − 3 x − 2 ) − ( x − 3 ) = − 4 ( 4 x + 5 ) + 1 3 ⇒ − 1 5 x − 1 0 − x + 3 = − 1 6 x − 2 0 + 1 3 ⇒ − 1 5 x − 1 0 − x + 3 = − 1 6 x − 2 0 + 1 3 ⇒ − 7 = − 7 → t r u e
Since the variable term − 1 6 x gets cancelled on both sides , the equation no more remains dependent on the value of x and thus any real value can satisfy the given equation.