Solve D=2/9(F+G)

$$D = \frac { 2 } { 9 } ( F + G )$$

Solve for F

$$F=\frac{9D}{2}-G$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{2}{9} by F+G.

$$D=\frac{2}{9}F+\frac{2}{9}G$$

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

$$\frac{2}{9}F+\frac{2}{9}G=D$$

Subtract \frac{2}{9}G from both sides.

$$\frac{2}{9}F=D-\frac{2}{9}G$$

The equation is in standard form.

$$\frac{2}{9}F=-\frac{2G}{9}+D$$

Divide both sides of the equation by \frac{2}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2}{9}F}{\frac{2}{9}}=\frac{-\frac{2G}{9}+D}{\frac{2}{9}}$$

Dividing by \frac{2}{9} undoes the multiplication by \frac{2}{9}.

$$F=\frac{-\frac{2G}{9}+D}{\frac{2}{9}}$$

Divide D-\frac{2G}{9} by \frac{2}{9} by multiplying D-\frac{2G}{9} by the reciprocal of \frac{2}{9}.

$$F=\frac{9D}{2}-G$$

Solve for D

$$D=\frac{2\left(F+G\right)}{9}$$

Solution Steps

Use the distributive property to multiply \frac{2}{9} by F+G.

$$D=\frac{2}{9}F+\frac{2}{9}G$$