11(k-2)+2[k-3(k+1)]=0

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Solution for 11(k-2)+2[k-3(k+1)]=0 equation:


Simplifying
11(k + -2) + 2[k + -3(k + 1)] = 0

Reorder the terms:
11(-2 + k) + 2[k + -3(k + 1)] = 0
(-2 * 11 + k * 11) + 2[k + -3(k + 1)] = 0
(-22 + 11k) + 2[k + -3(k + 1)] = 0

Reorder the terms:
-22 + 11k + 2[k + -3(1 + k)] = 0
-22 + 11k + 2[k + (1 * -3 + k * -3)] = 0
-22 + 11k + 2[k + (-3 + -3k)] = 0

Reorder the terms:
-22 + 11k + 2[-3 + k + -3k] = 0

Combine like terms: k + -3k = -2k
-22 + 11k + 2[-3 + -2k] = 0
-22 + 11k + [-3 * 2 + -2k * 2] = 0
-22 + 11k + [-6 + -4k] = 0

Reorder the terms:
-22 + -6 + 11k + -4k = 0

Combine like terms: -22 + -6 = -28
-28 + 11k + -4k = 0

Combine like terms: 11k + -4k = 7k
-28 + 7k = 0

Solving
-28 + 7k = 0

Solving for variable 'k'.

Move all terms containing k to the left, all other terms to the right.

Add '28' to each side of the equation.
-28 + 28 + 7k = 0 + 28

Combine like terms: -28 + 28 = 0
0 + 7k = 0 + 28
7k = 0 + 28

Combine like terms: 0 + 28 = 28
7k = 28

Divide each side by '7'.
k = 4

Simplifying
k = 4

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