4(n-3)-2(n+2)=3(n+5)-n

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Solution for 4(n-3)-2(n+2)=3(n+5)-n equation: 


Simplifying
4(n + -3) + -2(n + 2) = 3(n + 5) + -1n

Reorder the terms:
4(-3 + n) + -2(n + 2) = 3(n + 5) + -1n
(-3 * 4 + n * 4) + -2(n + 2) = 3(n + 5) + -1n
(-12 + 4n) + -2(n + 2) = 3(n + 5) + -1n

Reorder the terms:
-12 + 4n + -2(2 + n) = 3(n + 5) + -1n
-12 + 4n + (2 * -2 + n * -2) = 3(n + 5) + -1n
-12 + 4n + (-4 + -2n) = 3(n + 5) + -1n

Reorder the terms:
-12 + -4 + 4n + -2n = 3(n + 5) + -1n

Combine like terms: -12 + -4 = -16
-16 + 4n + -2n = 3(n + 5) + -1n

Combine like terms: 4n + -2n = 2n
-16 + 2n = 3(n + 5) + -1n

Reorder the terms:
-16 + 2n = 3(5 + n) + -1n
-16 + 2n = (5 * 3 + n * 3) + -1n
-16 + 2n = (15 + 3n) + -1n

Combine like terms: 3n + -1n = 2n
-16 + 2n = 15 + 2n

Add '-2n' to each side of the equation.
-16 + 2n + -2n = 15 + 2n + -2n

Combine like terms: 2n + -2n = 0
-16 + 0 = 15 + 2n + -2n
-16 = 15 + 2n + -2n

Combine like terms: 2n + -2n = 0
-16 = 15 + 0
-16 = 15

Solving
-16 = 15

Couldn't find a variable to solve for.

This equation is invalid, the left and right sides are not equal, therefore there is no solution.

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