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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