2(b+3c-5)-(b-2c)=

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Solution for 2(b+3c-5)-(b-2c)= equation:


Simplifying
2(b + 3c + -5) + -1(b + -2c) = 0

Reorder the terms:
2(-5 + b + 3c) + -1(b + -2c) = 0
(-5 * 2 + b * 2 + 3c * 2) + -1(b + -2c) = 0
(-10 + 2b + 6c) + -1(b + -2c) = 0
-10 + 2b + 6c + (b * -1 + -2c * -1) = 0
-10 + 2b + 6c + (-1b + 2c) = 0

Reorder the terms:
-10 + 2b + -1b + 6c + 2c = 0

Combine like terms: 2b + -1b = 1b
-10 + 1b + 6c + 2c = 0

Combine like terms: 6c + 2c = 8c
-10 + 1b + 8c = 0

Solving
-10 + 1b + 8c = 0

Solving for variable 'b'.

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

Add '10' to each side of the equation.
-10 + 1b + 10 + 8c = 0 + 10

Reorder the terms:
-10 + 10 + 1b + 8c = 0 + 10

Combine like terms: -10 + 10 = 0
0 + 1b + 8c = 0 + 10
1b + 8c = 0 + 10

Combine like terms: 0 + 10 = 10
1b + 8c = 10

Add '-8c' to each side of the equation.
1b + 8c + -8c = 10 + -8c

Combine like terms: 8c + -8c = 0
1b + 0 = 10 + -8c
1b = 10 + -8c

Divide each side by '1'.
b = 10 + -8c

Simplifying
b = 10 + -8c

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