Solve for d Hd+4Md=g^3

$H d + 4 M d = g^{3}$

Step 1

Factor $d$ out of $H d + 4 M d$ .

Factor $d$ out of $H d$ .

$d H + 4 M d = g^{3}$

Factor $d$ out of $4 M d$ .

$d H + d (4 M) = g^{3}$

Factor $d$ out of $d H + d (4 M)$ .

$d (H + 4 M) = g^{3}$

Step 2

Divide each term in $d (H + 4 M) = g^{3}$ by $H + 4 M$ and simplify.

Divide each term in $d (H + 4 M) = g^{3}$ by $H + 4 M$ .

$\frac{d (H + 4 M)}{H + 4 M} = \frac{g^{3}}{H + 4 M}$

Simplify the left side.

Cancel the common factor of $H + 4 M$ .

Cancel the common factor.

$\frac{d (H + 4 M)}{H + 4 M} = \frac{g^{3}}{H + 4 M}$

Divide $d$ by $1$ .

$d = \frac{g^{3}}{H + 4 M}$