What is the answer to solvefor b,gb+Nb=2F^2 ?

The answer to solvefor b,gb+Nb=2F^2 is b=(2F^2)/(g+N);\quad g\ne-N

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solvefor b,gb+Nb=2F^2

solve for

b, g b + N b = 2 F^{2}

b = \frac{2 F ^{2}}{g + N}; g \neq = - N

Solution steps

g b + N b = 2 F^{2}

Factor

g b + N b : b (g + N)

b (g + N) = 2 F^{2}

Divide both sides by

g + N; g \neq = - N

b = \frac{2 F ^{2}}{g + N}; g \neq = - N

5 (2 p + 4) = 2 (5 - 2 p)

- 4 + 5 x - 7 = 10 + 3 x - 2 x

- 4 x = - x + 3

x, ((y - 1)) \frac{2}{4} - ((x - 3)) \frac{2}{9} = 1

x, 4 x - 4 y^{3} = 0

What is the answer to solvefor b,gb+Nb=2F^2 ?
The answer to solvefor b,gb+Nb=2F^2 is b=(2F^2)/(g+N);\quad g\ne-N