3(n-2)+4(n+3)+2(n+4)=140

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Solution for 3(n-2)+4(n+3)+2(n+4)=140 equation:


Simplifying
3(n + -2) + 4(n + 3) + 2(n + 4) = 140

Reorder the terms:
3(-2 + n) + 4(n + 3) + 2(n + 4) = 140
(-2 * 3 + n * 3) + 4(n + 3) + 2(n + 4) = 140
(-6 + 3n) + 4(n + 3) + 2(n + 4) = 140

Reorder the terms:
-6 + 3n + 4(3 + n) + 2(n + 4) = 140
-6 + 3n + (3 * 4 + n * 4) + 2(n + 4) = 140
-6 + 3n + (12 + 4n) + 2(n + 4) = 140

Reorder the terms:
-6 + 3n + 12 + 4n + 2(4 + n) = 140
-6 + 3n + 12 + 4n + (4 * 2 + n * 2) = 140
-6 + 3n + 12 + 4n + (8 + 2n) = 140

Reorder the terms:
-6 + 12 + 8 + 3n + 4n + 2n = 140

Combine like terms: -6 + 12 = 6
6 + 8 + 3n + 4n + 2n = 140

Combine like terms: 6 + 8 = 14
14 + 3n + 4n + 2n = 140

Combine like terms: 3n + 4n = 7n
14 + 7n + 2n = 140

Combine like terms: 7n + 2n = 9n
14 + 9n = 140

Solving
14 + 9n = 140

Solving for variable 'n'.

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

Add '-14' to each side of the equation.
14 + -14 + 9n = 140 + -14

Combine like terms: 14 + -14 = 0
0 + 9n = 140 + -14
9n = 140 + -14

Combine like terms: 140 + -14 = 126
9n = 126

Divide each side by '9'.
n = 14

Simplifying
n = 14

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