Solve 1/2(b+c)+1000quadb=1/4(a+c)+2a

$$\frac { 1 } { 2 } ( b + c ) + 1000 \quad b = \frac { 1 } { 4 } ( a + c ) + 2 a$$

Solve for a

$$a=\frac{4002b+c}{9}$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{2} by b+c.

$$\frac{1}{2}b+\frac{1}{2}c+1000b=\frac{1}{4}\left(a+c\right)+2a$$

Combine \frac{1}{2}b and 1000b to get \frac{2001}{2}b.

$$\frac{2001}{2}b+\frac{1}{2}c=\frac{1}{4}\left(a+c\right)+2a$$

Use the distributive property to multiply \frac{1}{4} by a+c.

$$\frac{2001}{2}b+\frac{1}{2}c=\frac{1}{4}a+\frac{1}{4}c+2a$$

Combine \frac{1}{4}a and 2a to get \frac{9}{4}a.

$$\frac{2001}{2}b+\frac{1}{2}c=\frac{9}{4}a+\frac{1}{4}c$$

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

$$\frac{9}{4}a+\frac{1}{4}c=\frac{2001}{2}b+\frac{1}{2}c$$

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

$$\frac{9}{4}a=\frac{2001}{2}b+\frac{1}{2}c-\frac{1}{4}c$$

Combine \frac{1}{2}c and -\frac{1}{4}c to get \frac{1}{4}c.

$$\frac{9}{4}a=\frac{2001}{2}b+\frac{1}{4}c$$

The equation is in standard form.

$$\frac{9}{4}a=\frac{c}{4}+\frac{2001b}{2}$$

Divide both sides of the equation by \frac{9}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{9}{4}a}{\frac{9}{4}}=\frac{\frac{c}{4}+\frac{2001b}{2}}{\frac{9}{4}}$$

Dividing by \frac{9}{4} undoes the multiplication by \frac{9}{4}.

$$a=\frac{\frac{c}{4}+\frac{2001b}{2}}{\frac{9}{4}}$$

Divide \frac{2001b}{2}+\frac{c}{4} by \frac{9}{4} by multiplying \frac{2001b}{2}+\frac{c}{4} by the reciprocal of \frac{9}{4}.

$$a=\frac{c}{9}+\frac{1334b}{3}$$

Solve for b

$$b=\frac{3a}{1334}-\frac{c}{4002}$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{2} by b+c.

$$\frac{1}{2}b+\frac{1}{2}c+1000b=\frac{1}{4}\left(a+c\right)+2a$$

Combine \frac{1}{2}b and 1000b to get \frac{2001}{2}b.

$$\frac{2001}{2}b+\frac{1}{2}c=\frac{1}{4}\left(a+c\right)+2a$$

Use the distributive property to multiply \frac{1}{4} by a+c.

$$\frac{2001}{2}b+\frac{1}{2}c=\frac{1}{4}a+\frac{1}{4}c+2a$$

Combine \frac{1}{4}a and 2a to get \frac{9}{4}a.

$$\frac{2001}{2}b+\frac{1}{2}c=\frac{9}{4}a+\frac{1}{4}c$$

Subtract \frac{1}{2}c from both sides.

$$\frac{2001}{2}b=\frac{9}{4}a+\frac{1}{4}c-\frac{1}{2}c$$

Combine \frac{1}{4}c and -\frac{1}{2}c to get -\frac{1}{4}c.

$$\frac{2001}{2}b=\frac{9}{4}a-\frac{1}{4}c$$

The equation is in standard form.

$$\frac{2001}{2}b=\frac{9a-c}{4}$$

Divide both sides of the equation by \frac{2001}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2001}{2}b}{\frac{2001}{2}}=\frac{9a-c}{4\times \frac{2001}{2}}$$

Dividing by \frac{2001}{2} undoes the multiplication by \frac{2001}{2}.

$$b=\frac{9a-c}{4\times \frac{2001}{2}}$$

Divide \frac{9a-c}{4} by \frac{2001}{2} by multiplying \frac{9a-c}{4} by the reciprocal of \frac{2001}{2}.

$$b=\frac{3a}{1334}-\frac{c}{4002}$$