Solve for b

b=8+\frac{12}{x},x\neq 0

Solve for x

x=-\frac{12}{8-b},b\neq 8

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bx-7=8x+5

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

bx=8x+5+7

Add 7 to both sides.

bx=8x+12

Add 5 and 7 to get 12.

xb=8x+12

The equation is in standard form.

\frac{xb}{x}=\frac{8x+12}{x}

Divide both sides by x.

b=\frac{8x+12}{x}

Dividing by x undoes the multiplication by x.

b=8+\frac{12}{x}

Divide 8x+12 by x.

8x+5-bx=-7

Subtract bx from both sides.

8x-bx=-7-5

Subtract 5 from both sides.

8x-bx=-12

Subtract 5 from -7 to get -12.

\left(8-b\right)x=-12

Combine all terms containing x.

\frac{\left(8-b\right)x}{8-b}=\frac{-12}{8-b}

Divide both sides by 8-b.

x=\frac{-12}{8-b}

Dividing by 8-b undoes the multiplication by 8-b.

x=-\frac{12}{8-b}

Divide -12 by 8-b.

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