Solve 2left(a+b+cright)=(a+b+c)k

$$2 \left( a+b+c \right) =(a+b+c)k$$

Solve for a

$$\left\{\begin{matrix}\\a=-\left(b+c\right)\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&k=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by a+b+c.

$$2a+2b+2c=\left(a+b+c\right)k$$

Use the distributive property to multiply a+b+c by k.

$$2a+2b+2c=ak+bk+ck$$

Subtract ak from both sides.

$$2a+2b+2c-ak=bk+ck$$

Subtract 2b from both sides.

$$2a+2c-ak=bk+ck-2b$$

Subtract 2c from both sides.

$$2a-ak=bk+ck-2b-2c$$

Combine all terms containing a.

$$\left(2-k\right)a=bk+ck-2b-2c$$

The equation is in standard form.

$$\left(2-k\right)a=bk-2b+ck-2c$$

Divide both sides by 2-k.

$$\frac{\left(2-k\right)a}{2-k}=\frac{\left(k-2\right)\left(b+c\right)}{2-k}$$

Dividing by 2-k undoes the multiplication by 2-k.

$$a=\frac{\left(k-2\right)\left(b+c\right)}{2-k}$$

Divide \left(-2+k\right)\left(b+c\right) by 2-k.

$$a=-\left(b+c\right)$$

Solve for b

$$\left\{\begin{matrix}\\b=-\left(a+c\right)\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&k=2\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by a+b+c.

$$2a+2b+2c=\left(a+b+c\right)k$$

Use the distributive property to multiply a+b+c by k.

$$2a+2b+2c=ak+bk+ck$$

Subtract bk from both sides.

$$2a+2b+2c-bk=ak+ck$$

Subtract 2a from both sides.

$$2b+2c-bk=ak+ck-2a$$

Subtract 2c from both sides.

$$2b-bk=ak+ck-2a-2c$$

Combine all terms containing b.

$$\left(2-k\right)b=ak+ck-2a-2c$$

The equation is in standard form.

$$\left(2-k\right)b=ak-2a+ck-2c$$

Divide both sides by 2-k.

$$\frac{\left(2-k\right)b}{2-k}=\frac{\left(k-2\right)\left(a+c\right)}{2-k}$$

Dividing by 2-k undoes the multiplication by 2-k.

$$b=\frac{\left(k-2\right)\left(a+c\right)}{2-k}$$

Divide \left(-2+k\right)\left(a+c\right) by 2-k.

$$b=-\left(a+c\right)$$