How do you solve the equation and identify any extraneous solutions for #(x4)/x + x/3=6#?
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Solve using the quadratic formula:
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To solve the equation (x4)/x + x/3 = 6 and identify any extraneous solutions, we can follow these steps:

Multiply every term in the equation by the common denominator, which is 3x. This eliminates the fractions. (3x)((x4)/x) + (3x)(x/3) = (3x)(6)

Simplify each term by canceling out common factors. 3(x4) + x(x) = 18x

Distribute and combine like terms. 3x  12 + x^2 = 18x

Rearrange the equation to bring all terms to one side and set it equal to zero. x^2 + 15x  12 = 0

Factor the quadratic equation or use the quadratic formula to solve for x. (x + 16)(x  3) = 0 x = 16 or x = 3

Check each solution by substituting it back into the original equation. For x = 16: (164)/(16) + (16)/3 = 6 For x = 3: (34)/(3) + (3)/3 = 6

Identify any extraneous solutions by checking if they make any denominators zero or result in undefined values. None of the solutions make any denominators zero or result in undefined values.
Therefore, the solutions to the equation are x = 16 and x = 3, and there are no extraneous solutions.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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