Solve for b

b=-5+\frac{1}{3x},x\neq 0

$b=−5+3x1 ,x=0$

Solve for x

x=\frac{1}{3\left(b+5\right)},b\neq -5

$x=3(b+5)1 ,b=−5$

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bx+\frac{1}{3}=\frac{2}{3}-5x

Swap sides so that all variable terms are on the left hand side.

bx=\frac{2}{3}-5x-\frac{1}{3}

Subtract \frac{1}{3} from both sides.

bx=\frac{1}{3}-5x

Subtract \frac{1}{3} from \frac{2}{3} to get \frac{1}{3}.

xb=\frac{1}{3}-5x

The equation is in standard form.

\frac{xb}{x}=\frac{\frac{1}{3}-5x}{x}

Divide both sides by x.

b=\frac{\frac{1}{3}-5x}{x}

Dividing by x undoes the multiplication by x.

b=-5+\frac{1}{3x}

Divide \frac{1}{3}-5x by x.

\frac{2}{3}-5x-bx=\frac{1}{3}

Subtract bx from both sides.

-5x-bx=\frac{1}{3}-\frac{2}{3}

Subtract \frac{2}{3} from both sides.

-5x-bx=-\frac{1}{3}

Subtract \frac{2}{3} from \frac{1}{3} to get -\frac{1}{3}.

\left(-5-b\right)x=-\frac{1}{3}

Combine all terms containing x.

\left(-b-5\right)x=-\frac{1}{3}

The equation is in standard form.

\frac{\left(-b-5\right)x}{-b-5}=\frac{-\frac{1}{3}}{-b-5}

Divide both sides by -5-b.

x=\frac{-\frac{1}{3}}{-b-5}

Dividing by -5-b undoes the multiplication by -5-b.

x=\frac{1}{3\left(b+5\right)}

Divide -\frac{1}{3} by -5-b.

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