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How do you solve #(5k + 5)/(3k^2) = (k - 4)/(3k^2) + 1/k^2# and find any extraneous solutions?

Answer 1

Avery Alexander

#k=-3/2#

bringing down to the same amount

#(5k+5)/(3k^2)=(k-4)/(3k^2)+3/(3k^2)#

#(5k+5)/(3k^2)-(k-4)/(3k^2)-3/(3k^2)=0#

#((5k+5)-(k-4)-3)/(3k^2)=0#

#(4k+6)/(3k^2)=0#

Solving for #k# we obtain #k=-3/2#

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Answer 2

Julia Adams

To solve the equation (5k + 5)/(3k^2) = (k - 4)/(3k^2) + 1/k^2 and find any extraneous solutions, we can follow these steps:

Multiply both sides of the equation by 3k^2 to eliminate the denominators: (5k + 5) = (k - 4) + 3k^2/k^2
Simplify the equation: 5k + 5 = k - 4 + 3
Combine like terms: 5k + 5 = k - 1 + 3
Simplify further: 5k + 5 = k + 2
Subtract k from both sides: 4k + 5 = 2
Subtract 5 from both sides: 4k = -3
Divide both sides by 4: k = -3/4

Therefore, the solution to the equation is k = -3/4. To check for extraneous solutions, substitute this value back into the original equation and verify if it holds true.

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Answer from HIX Tutor

When evaluating a one-sided 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.