Solve b=4a-2b+1/4-3a

$$b = \frac { 4 a - 2 b + 1 } { 4 } - 3 a$$

Solve for a

$$a=-\frac{3b}{4}+\frac{1}{8}$$

Steps for Solving Linear Equation

Divide each term of 4a-2b+1 by 4 to get a-\frac{1}{2}b+\frac{1}{4}.

$$b=a-\frac{1}{2}b+\frac{1}{4}-3a$$

Combine a and -3a to get -2a.

$$b=-2a-\frac{1}{2}b+\frac{1}{4}$$

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

$$-2a-\frac{1}{2}b+\frac{1}{4}=b$$

Add \frac{1}{2}b to both sides.

$$-2a+\frac{1}{4}=b+\frac{1}{2}b$$

Combine b and \frac{1}{2}b to get \frac{3}{2}b.

$$-2a+\frac{1}{4}=\frac{3}{2}b$$

Subtract \frac{1}{4} from both sides.

$$-2a=\frac{3}{2}b-\frac{1}{4}$$

The equation is in standard form.

$$-2a=\frac{3b}{2}-\frac{1}{4}$$

Divide both sides by -2.

$$\frac{-2a}{-2}=\frac{\frac{3b}{2}-\frac{1}{4}}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$a=\frac{\frac{3b}{2}-\frac{1}{4}}{-2}$$

Divide \frac{3b}{2}-\frac{1}{4} by -2.

$$a=-\frac{3b}{4}+\frac{1}{8}$$

Solve for b

$$b=-\frac{4a}{3}+\frac{1}{6}$$

Steps for Solving Linear Equation

Divide each term of 4a-2b+1 by 4 to get a-\frac{1}{2}b+\frac{1}{4}.

$$b=a-\frac{1}{2}b+\frac{1}{4}-3a$$

Combine a and -3a to get -2a.

$$b=-2a-\frac{1}{2}b+\frac{1}{4}$$

Add \frac{1}{2}b to both sides.

$$b+\frac{1}{2}b=-2a+\frac{1}{4}$$

Combine b and \frac{1}{2}b to get \frac{3}{2}b.

$$\frac{3}{2}b=-2a+\frac{1}{4}$$

The equation is in standard form.

$$\frac{3}{2}b=\frac{1}{4}-2a$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{2}b}{\frac{3}{2}}=\frac{\frac{1}{4}-2a}{\frac{3}{2}}$$

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

$$b=\frac{\frac{1}{4}-2a}{\frac{3}{2}}$$

Divide -2a+\frac{1}{4} by \frac{3}{2} by multiplying -2a+\frac{1}{4} by the reciprocal of \frac{3}{2}.

$$b=-\frac{4a}{3}+\frac{1}{6}$$