(c^2+3c)-(c^2-4c)=9c-16

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Solution for (c^2+3c)-(c^2-4c)=9c-16 equation:


Simplifying
(c2 + 3c) + -1(c2 + -4c) = 9c + -16

Reorder the terms:
(3c + c2) + -1(c2 + -4c) = 9c + -16

Remove parenthesis around (3c + c2)
3c + c2 + -1(c2 + -4c) = 9c + -16

Reorder the terms:
3c + c2 + -1(-4c + c2) = 9c + -16
3c + c2 + (-4c * -1 + c2 * -1) = 9c + -16
3c + c2 + (4c + -1c2) = 9c + -16

Reorder the terms:
3c + 4c + c2 + -1c2 = 9c + -16

Combine like terms: 3c + 4c = 7c
7c + c2 + -1c2 = 9c + -16

Combine like terms: c2 + -1c2 = 0
7c + 0 = 9c + -16
7c = 9c + -16

Reorder the terms:
7c = -16 + 9c

Solving
7c = -16 + 9c

Solving for variable 'c'.

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

Add '-9c' to each side of the equation.
7c + -9c = -16 + 9c + -9c

Combine like terms: 7c + -9c = -2c
-2c = -16 + 9c + -9c

Combine like terms: 9c + -9c = 0
-2c = -16 + 0
-2c = -16

Divide each side by '-2'.
c = 8

Simplifying
c = 8

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