Solve {l}{tx+y=2sqrt{t}}{(t-1)^2/4tx+y=(t-1)/sqrt{t}}

$$\left\{ \begin{array} { l } { t x + y = 2 \sqrt { t } } \\ { \frac { ( t - 1 ) ^ { 2 } } { 4 t } x + y = \frac { ( t - 1 ) } { \sqrt { t } } } \end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{4\sqrt{t}}{3t-1}\text{, }y=\frac{2\sqrt{t}\left(t-1\right)}{3t-1}\text{, }&t\neq \frac{1}{3}\text{ and }t\neq -1\text{ and }t\neq 0\\x\in \mathrm{C}\text{, }y=x+2i\text{, }&t=-1\end{matrix}\right.$$

Steps Using Substitution

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.

$$tx+y=2\sqrt{t},\frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$tx+y=2\sqrt{t}$$

Subtract y from both sides of the equation.

$$tx=-y+2\sqrt{t}$$

Divide both sides by t.

$$x=\frac{1}{t}\left(-y+2\sqrt{t}\right)$$

Multiply \frac{1}{t} times -y+2\sqrt{t}.

$$x=\left(-\frac{1}{t}\right)y+2t^{-\frac{1}{2}}$$

Substitute \frac{2\sqrt{t}-y}{t} for x in the other equation, \frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}.

$$\frac{\left(t-1\right)^{2}}{4t}\left(\left(-\frac{1}{t}\right)y+2t^{-\frac{1}{2}}\right)+y=\frac{t-1}{\sqrt{t}}$$

Multiply \frac{1}{4}\left(t-1\right)^{2}t^{-1} times \frac{2\sqrt{t}-y}{t}.

$$\left(-\frac{\left(t-1\right)^{2}}{4t^{2}}\right)y+\frac{t^{-\frac{3}{2}}\left(t-1\right)^{2}}{2}+y=\frac{t-1}{\sqrt{t}}$$

Add -\frac{\left(t-1\right)^{2}y}{4t^{2}} to y.

$$\left(\frac{3}{4}+\frac{\frac{t}{2}-\frac{1}{4}}{t^{2}}\right)y+\frac{t^{-\frac{3}{2}}\left(t-1\right)^{2}}{2}=\frac{t-1}{\sqrt{t}}$$

Subtract \frac{\left(t-1\right)^{2}t^{-\frac{3}{2}}}{2} from both sides of the equation.

$$\left(\frac{3}{4}+\frac{\frac{t}{2}-\frac{1}{4}}{t^{2}}\right)y=\frac{\sqrt{t}-t^{-\frac{3}{2}}}{2}$$

Divide both sides by \frac{3}{4}+\frac{-\frac{1}{4}+\frac{t}{2}}{t^{2}}.

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

Substitute \frac{2\sqrt{t}\left(t-1\right)}{-1+3t} for y in x=\left(-\frac{1}{t}\right)y+2t^{-\frac{1}{2}}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\left(-\frac{1}{t}\right)\times \frac{2\sqrt{t}\left(t-1\right)}{3t-1}+2t^{-\frac{1}{2}}$$

Multiply -\frac{1}{t} times \frac{2\sqrt{t}\left(t-1\right)}{-1+3t}.

$$x=-\frac{2t^{-\frac{1}{2}}\left(t-1\right)}{3t-1}+2t^{-\frac{1}{2}}$$

Add 2t^{-\frac{1}{2}} to -\frac{2\left(t-1\right)t^{-\frac{1}{2}}}{-1+3t}.

$$x=\frac{4\sqrt{t}}{3t-1}$$

The system is now solved.

$$x=\frac{4\sqrt{t}}{3t-1},y=\frac{2\sqrt{t}\left(t-1\right)}{3t-1}$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$tx+y=2\sqrt{t},\frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}$$

Subtract \frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}} from tx+y=2\sqrt{t} by subtracting like terms on each side of the equal sign.

$$tx+\left(-\frac{\left(t-1\right)^{2}}{4t}\right)x+y-y=2\sqrt{t}-t^{-\frac{1}{2}}\left(t-1\right)$$

Add y to -y. Terms y and -y cancel out, leaving an equation with only one variable that can be solved.

$$tx+\left(-\frac{\left(t-1\right)^{2}}{4t}\right)x=2\sqrt{t}-t^{-\frac{1}{2}}\left(t-1\right)$$

Add tx to -\frac{\left(t-1\right)^{2}x}{4t}.

$$\left(\frac{3t}{4}+\frac{1}{2}-\frac{1}{4t}\right)x=2\sqrt{t}-t^{-\frac{1}{2}}\left(t-1\right)$$

Add 2\sqrt{t} to -\left(t-1\right)t^{-\frac{1}{2}}.

$$\left(\frac{3t}{4}+\frac{1}{2}-\frac{1}{4t}\right)x=t^{-\frac{1}{2}}\left(t+1\right)$$

Divide both sides by -\frac{1}{4t}+\frac{1}{2}+\frac{3t}{4}.

$$x=\frac{4\sqrt{t}}{3t-1}$$

Substitute \frac{4\sqrt{t}}{-1+3t} for x in \frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}. Because the resulting equation contains only one variable, you can solve for y directly.

$$\frac{\left(t-1\right)^{2}}{4t}\times \frac{4\sqrt{t}}{3t-1}+y=\frac{t-1}{\sqrt{t}}$$

Multiply \frac{1}{4}\left(t-1\right)^{2}t^{-1} times \frac{4\sqrt{t}}{-1+3t}.

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

Subtract \frac{t^{-\frac{1}{2}}\left(t-1\right)^{2}}{-1+3t} from both sides of the equation.

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

The system is now solved.

$$x=\frac{4\sqrt{t}}{3t-1},y=\frac{2\sqrt{t}\left(t-1\right)}{3t-1}$$

Solve for x, y

$$x=\frac{4\sqrt{t}}{3t-1}$$

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

$$t\neq \frac{1}{3}\text{ and }t>0$$

Steps Using Substitution

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.

$$tx+y=2\sqrt{t},\frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$tx+y=2\sqrt{t}$$

Subtract y from both sides of the equation.

$$tx=-y+2\sqrt{t}$$

Divide both sides by t.

$$x=\frac{1}{t}\left(-y+2\sqrt{t}\right)$$

Multiply \frac{1}{t} times -y+2\sqrt{t}.

$$x=\left(-\frac{1}{t}\right)y+\frac{2}{\sqrt{t}}$$

Substitute \frac{2\sqrt{t}-y}{t} for x in the other equation, \frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}.

$$\frac{\left(t-1\right)^{2}}{4t}\left(\left(-\frac{1}{t}\right)y+\frac{2}{\sqrt{t}}\right)+y=\frac{t-1}{\sqrt{t}}$$

Multiply \frac{1}{4}\left(t-1\right)^{2}t^{-1} times \frac{2\sqrt{t}-y}{t}.

$$\left(-\frac{\left(t-1\right)^{2}}{4t^{2}}\right)y+\frac{\left(t-1\right)^{2}}{2t^{\frac{3}{2}}}+y=\frac{t-1}{\sqrt{t}}$$

Add -\frac{\left(t-1\right)^{2}y}{4t^{2}} to y.

$$\left(\frac{3}{4}+\frac{\frac{t}{2}-\frac{1}{4}}{t^{2}}\right)y+\frac{\left(t-1\right)^{2}}{2t^{\frac{3}{2}}}=\frac{t-1}{\sqrt{t}}$$

Subtract \frac{\left(t-1\right)^{2}}{2t^{\frac{3}{2}}} from both sides of the equation.

$$\left(\frac{3}{4}+\frac{\frac{t}{2}-\frac{1}{4}}{t^{2}}\right)y=\frac{\sqrt{t}}{2}-\frac{1}{2t^{\frac{3}{2}}}$$

Divide both sides by \frac{3}{4}+\frac{-\frac{1}{4}+\frac{t}{2}}{t^{2}}.

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

Substitute \frac{2\sqrt{t}\left(-1+t\right)}{-1+3t} for y in x=\left(-\frac{1}{t}\right)y+\frac{2}{\sqrt{t}}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\left(-\frac{1}{t}\right)\times \frac{2\sqrt{t}\left(t-1\right)}{3t-1}+\frac{2}{\sqrt{t}}$$

Multiply -\frac{1}{t} times \frac{2\sqrt{t}\left(-1+t\right)}{-1+3t}.

$$x=-\frac{2\left(t-1\right)}{\sqrt{t}\left(3t-1\right)}+\frac{2}{\sqrt{t}}$$

Add \frac{2}{\sqrt{t}} to -\frac{2\left(-1+t\right)}{\left(-1+3t\right)\sqrt{t}}.

$$x=\frac{4\sqrt{t}}{3t-1}$$

The system is now solved.

$$x=\frac{4\sqrt{t}}{3t-1},y=\frac{2\sqrt{t}\left(t-1\right)}{3t-1}$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$tx+y=2\sqrt{t},\frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}$$

Subtract \frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}} from tx+y=2\sqrt{t} by subtracting like terms on each side of the equal sign.

$$tx+\left(-\frac{\left(t-1\right)^{2}}{4t}\right)x+y-y=2\sqrt{t}-\frac{t-1}{\sqrt{t}}$$

Add y to -y. Terms y and -y cancel out, leaving an equation with only one variable that can be solved.

$$tx+\left(-\frac{\left(t-1\right)^{2}}{4t}\right)x=2\sqrt{t}-\frac{t-1}{\sqrt{t}}$$

Add tx to -\frac{\left(t-1\right)^{2}x}{4t}.

$$\left(\frac{3t}{4}+\frac{1}{2}-\frac{1}{4t}\right)x=2\sqrt{t}-\frac{t-1}{\sqrt{t}}$$

Add 2\sqrt{t} to -\frac{t-1}{\sqrt{t}}.

$$\left(\frac{3t}{4}+\frac{1}{2}-\frac{1}{4t}\right)x=\frac{t+1}{\sqrt{t}}$$

Divide both sides by -\frac{1}{4t}+\frac{1}{2}+\frac{3t}{4}.

$$x=\frac{4\sqrt{t}}{3t-1}$$

Substitute \frac{4\sqrt{t}}{-1+3t} for x in \frac{\left(t-1\right)^{2}}{4t}x+y=\frac{t-1}{\sqrt{t}}. Because the resulting equation contains only one variable, you can solve for y directly.

$$\frac{\left(t-1\right)^{2}}{4t}\times \frac{4\sqrt{t}}{3t-1}+y=\frac{t-1}{\sqrt{t}}$$

Multiply \frac{1}{4}\left(t-1\right)^{2}t^{-1} times \frac{4\sqrt{t}}{-1+3t}.

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

Subtract \frac{\left(t-1\right)^{2}}{\left(-1+3t\right)\sqrt{t}} from both sides of the equation.

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

The system is now solved.

$$x=\frac{4\sqrt{t}}{3t-1},y=\frac{2\sqrt{t}\left(t-1\right)}{3t-1}$$