How do you solve #5+ 7k = k - 7#?

Answer 1

Answer: #k=-2#

Solve #5+7k=k-7#
First we want to isolate the variable #k# on one side of the equation by subtracting #k# and #5# from both sides: #5+7k-k-5=k-7-k-5#
We can simplify: #6k=-12#
Finally, we divide both sides by #6# to solve for #k#: #(6k)/6=-12/6#
#k=-2#
Substituting into the original equation to check: #5+7(-2)=(-2)-7#
#5-14=-9#
#-9=-9#, which is true
Therefore, our answer is #k=-2#
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Answer 2

To solve (5 + 7k = k - 7), follow these steps:

  1. Move all terms involving (k) to one side of the equation and constants to the other side: (5 + 7k - k = -7).

  2. Combine like terms: (5 + 6k = -7).

  3. Move the constant term to the other side of the equation: (6k = -7 - 5).

  4. Simplify: (6k = -12).

  5. Divide both sides by 6 to solve for (k): (k = \frac{-12}{6}).

  6. Simplify the expression: (k = -2).

Therefore, the solution to the equation (5 + 7k = k - 7) is (k = -2).

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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.

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.

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.

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.

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