Solve a(q-c)=dt

$$a ( q - c ) = d t$$

Solve for a

$$\left\{\begin{matrix}a=\frac{dt}{q-c}\text{, }&q\neq c\\a\in \mathrm{R}\text{, }&\left(t=0\text{ or }d=0\right)\text{ and }q=c\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by q-c.

$$aq-ac=dt$$

Combine all terms containing a.

$$\left(q-c\right)a=dt$$

Divide both sides by q-c.

$$\frac{\left(q-c\right)a}{q-c}=\frac{dt}{q-c}$$

Dividing by q-c undoes the multiplication by q-c.

$$a=\frac{dt}{q-c}$$

Solve for c

$$\left\{\begin{matrix}c=-\frac{dt}{a}+q\text{, }&a\neq 0\\c\in \mathrm{R}\text{, }&\left(d=0\text{ or }t=0\right)\text{ and }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a by q-c.

$$aq-ac=dt$$

Subtract aq from both sides.

$$-ac=dt-aq$$

The equation is in standard form.

$$\left(-a\right)c=dt-aq$$

Divide both sides by -a.

$$\frac{\left(-a\right)c}{-a}=\frac{dt-aq}{-a}$$

Dividing by -a undoes the multiplication by -a.

$$c=\frac{dt-aq}{-a}$$

Divide dt-aq by -a.

$$c=-\frac{dt}{a}+q$$