Solve 3a+7b+6c=12

$$3 a + 7 b + 6 c = 12$$

Solve for a

$$a=-\frac{7b}{3}-2c+4$$

Steps for Solving Linear Equation

Subtract 7b from both sides.

$$3a+6c=12-7b$$

Subtract 6c from both sides.

$$3a=12-7b-6c$$

The equation is in standard form.

$$3a=12-6c-7b$$

Divide both sides by 3.

$$\frac{3a}{3}=\frac{12-6c-7b}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a=\frac{12-6c-7b}{3}$$

Divide 12-7b-6c by 3.

$$a=-\frac{7b}{3}-2c+4$$

Solve for b

$$b=\frac{12-6c-3a}{7}$$

Steps for Solving Linear Equation

Subtract 3a from both sides.

$$7b+6c=12-3a$$

Subtract 6c from both sides.

$$7b=12-3a-6c$$

The equation is in standard form.

$$7b=12-6c-3a$$

Divide both sides by 7.

$$\frac{7b}{7}=\frac{12-6c-3a}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$b=\frac{12-6c-3a}{7}$$