How do you solve #k- 8- 3= - 6( k + 6) - ( - 5k + 5)#?

Answer 1

#k = -15#

First, expand the terms in parenthesis and group and combine like terms on each side of the equation:

#k+ (-8 - 3) = -6k - 36 + 5k - 5#
#k - 11 = -6x + 5k - 36 - 5#
#k - 11 = (-6 + 5)k - 41#
#k - 11 = -1k - 41#
Now, we can isolate the #k# terms on one side of the equation and the constants on the other while keeping the equation balanced:
#k + k - 11 + 11 = -k + k - 41 + 11#
#2k - 0 = 0 - 30#
#2k = -30#
Now we can solve for #k# while keeping the equation balanced:
#(2k)/2 = (-30)/2#
#(cancel(2)k)/cancel(2) = -15#
#k = -15#
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Answer 2

To solve the equation (k - 8 - 3 = -6(k + 6) - (-5k + 5)), follow these steps:

  1. First, simplify both sides of the equation by distributing the -6 and -1 through the parentheses: [k - 8 - 3 = -6k - 36 + 5k - 5]

  2. Combine like terms on each side of the equation: [k - 11 = -k - 41]

  3. Add (k) to both sides of the equation to isolate the variable: [2k - 11 = -41]

  4. Add 11 to both sides of the equation: [2k = -30]

  5. Finally, divide both sides of the equation by 2 to solve for (k): [k = -15]

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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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