Solve k(a+b)-1=c

$$k ( a + b ) - 1 = c$$

Solve for a

$$\left\{\begin{matrix}a=\frac{1+c-bk}{k}\text{, }&k\neq 0\\a\in \mathrm{R}\text{, }&c=-1\text{ and }k=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$ka+kb-1=c$$

Subtract kb from both sides.

$$ka-1=c-kb$$

Add 1 to both sides.

$$ka=c-kb+1$$

The equation is in standard form.

$$ka=1+c-bk$$

Divide both sides by k.

$$\frac{ka}{k}=\frac{1+c-bk}{k}$$

Dividing by k undoes the multiplication by k.

$$a=\frac{1+c-bk}{k}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{1+c-ak}{k}\text{, }&k\neq 0\\b\in \mathrm{R}\text{, }&c=-1\text{ and }k=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$ka+kb-1=c$$

Subtract ka from both sides.

$$kb-1=c-ka$$

Add 1 to both sides.

$$kb=c-ka+1$$

The equation is in standard form.

$$kb=1+c-ak$$

Divide both sides by k.

$$\frac{kb}{k}=\frac{1+c-ak}{k}$$

Dividing by k undoes the multiplication by k.

$$b=\frac{1+c-ak}{k}$$