d x + 8 = 8 x + 7
Solve for d
d=8-\frac{1}{x}
x\neq 0
Solve for x
x=-\frac{1}{d-8}
d\neq 8
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dx=8x+7-8
Subtract 8 from both sides.
dx=8x-1
Subtract 8 from 7 to get -1.
xd=8x-1
The equation is in standard form.
\frac{xd}{x}=\frac{8x-1}{x}
Divide both sides by x.
d=\frac{8x-1}{x}
Dividing by x undoes the multiplication by x.
d=8-\frac{1}{x}
Divide 8x-1 by x.
dx+8-8x=7
Subtract 8x from both sides.
dx-8x=7-8
Subtract 8 from both sides.
dx-8x=-1
Subtract 8 from 7 to get -1.
\left(d-8\right)x=-1
Combine all terms containing x.
\frac{\left(d-8\right)x}{d-8}=-\frac{1}{d-8}
Divide both sides by d-8.
x=-\frac{1}{d-8}
Dividing by d-8 undoes the multiplication by d-8.
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