Solve tdx=(t-2x)dt,tdy=(tx+ty+2x-t)dt

$$t d x = ( t - 2 x ) d t , t d y = ( t x + t y + 2 x - t ) d t$$

Solve for x, y

$$\left\{\begin{matrix}x=\frac{t}{3}\text{, }y=-\frac{t}{3}\text{, }&t\neq 1\\x=0\text{, }y\in \mathrm{R}\text{, }&t=0\\x=\frac{t}{3}\text{, }y\in \mathrm{R}\text{, }&d=0\text{ or }t=1\\x\in \mathrm{R}\text{, }y\in \mathrm{R}\text{, }&d=0\text{ or }t=0\end{matrix}\right.$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply t-2x by d.

$$tdx=\left(td-2xd\right)t$$

Use the distributive property to multiply td-2xd by t.

$$tdx=dt^{2}-2xdt$$

Add 2xdt to both sides.

$$tdx+2xdt=dt^{2}$$

Combine tdx and 2xdt to get 3tdx.

$$3tdx=dt^{2}$$

Consider the second equation. Use the distributive property to multiply tx+ty+2x-t by d.

$$tdy=\left(txd+tyd+2xd-td\right)t$$

Use the distributive property to multiply txd+tyd+2xd-td by t.

$$tdy=xdt^{2}+ydt^{2}+2xdt-dt^{2}$$

Subtract xdt^{2} from both sides.

$$tdy-xdt^{2}=ydt^{2}+2xdt-dt^{2}$$

Subtract ydt^{2} from both sides.

$$tdy-xdt^{2}-ydt^{2}=2xdt-dt^{2}$$

Subtract 2xdt from both sides.

$$tdy-xdt^{2}-ydt^{2}-2xdt=-dt^{2}$$

Reorder the terms.

$$-dxt^{2}-2dtx+dty-dyt^{2}=-dt^{2}$$

Combine all terms containing x,y.

$$\left(-dt^{2}-2dt\right)x+\left(dt-dt^{2}\right)y=-dt^{2}$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$3dtx=dt^{2},\left(-dt^{2}-2dt\right)x+\left(dt-dt^{2}\right)y=-dt^{2}$$

Pick one of the two equations which is more simple to solve for x by isolating x on the left hand side of the equal sign.

$$3dtx=dt^{2}$$

Divide both sides by 3td.

$$x=\frac{t}{3}$$

Substitute \frac{t}{3} for x in the other equation, \left(-dt^{2}-2dt\right)x+\left(dt-dt^{2}\right)y=-dt^{2}.

$$\left(-dt^{2}-2dt\right)\times \frac{t}{3}+\left(dt-dt^{2}\right)y=-dt^{2}$$

Multiply -dt^{2}-2dt times \frac{t}{3}.

$$-\frac{d\left(t+2\right)t^{2}}{3}+\left(dt-dt^{2}\right)y=-dt^{2}$$

Add \frac{d\left(2+t\right)t^{2}}{3} to both sides of the equation.

$$dt\left(1-t\right)y=\frac{d\left(t-1\right)t^{2}}{3}$$

Divide both sides by td\left(1-t\right).

$$y=-\frac{t}{3}$$

The system is now solved.

$$x=\frac{t}{3},y=-\frac{t}{3}$$