Solve 2ar+rb+ta+tb=5a+2b

$$2 a r + r b + t a + t b = 5 a + 2 b$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{b\left(r+t-2\right)}{2r+t-5}\text{, }&r\neq \frac{5-t}{2}\\a\in \mathrm{R}\text{, }&\left(r=3\text{ and }t=-1\right)\text{ or }\left(b=0\text{ and }r=\frac{5-t}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 5a from both sides.

$$2ar+rb+ta+tb-5a=2b$$

Subtract rb from both sides.

$$2ar+ta+tb-5a=2b-rb$$

Subtract tb from both sides.

$$2ar+ta-5a=2b-rb-tb$$

Combine all terms containing a.

$$\left(2r+t-5\right)a=2b-rb-tb$$

The equation is in standard form.

$$\left(2r+t-5\right)a=2b-bt-br$$

Divide both sides by 2r+t-5.

$$\frac{\left(2r+t-5\right)a}{2r+t-5}=\frac{b\left(2-t-r\right)}{2r+t-5}$$

Dividing by 2r+t-5 undoes the multiplication by 2r+t-5.

$$a=\frac{b\left(2-t-r\right)}{2r+t-5}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{a\left(2r+t-5\right)}{r+t-2}\text{, }&r\neq 2-t\\b\in \mathrm{R}\text{, }&\left(a=0\text{ and }r=2-t\right)\text{ or }\left(r=3\text{ and }t=-1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract 2b from both sides.

$$2ar+rb+ta+tb-2b=5a$$

Subtract 2ar from both sides.

$$rb+ta+tb-2b=5a-2ar$$

Subtract ta from both sides.

$$rb+tb-2b=5a-2ar-ta$$

Combine all terms containing b.

$$\left(r+t-2\right)b=5a-2ar-ta$$

The equation is in standard form.

$$\left(r+t-2\right)b=5a-at-2ar$$

Divide both sides by -2+r+t.

$$\frac{\left(r+t-2\right)b}{r+t-2}=\frac{a\left(5-t-2r\right)}{r+t-2}$$

Dividing by -2+r+t undoes the multiplication by -2+r+t.

$$b=\frac{a\left(5-t-2r\right)}{r+t-2}$$