Solve {l}{3x-y=sqrt{2/3}}{x+3y=sqrt{9}}

$$\left\{ \begin{array} { l } { 3 x - y = \sqrt { \frac { 2 } { 3 } } } \\ { x + 3 y = \sqrt { 9 } } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{6}+3}{10}\approx 0.544948974$$

$$y=-\frac{\sqrt{6}}{30}+\frac{9}{10}\approx 0.818350342$$

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.

$$3x-y=\sqrt{\frac{2}{3}},x+3y=\sqrt{9}$$

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

$$3x-y=\sqrt{\frac{2}{3}}$$

Add y to both sides of the equation.

$$3x=y+\frac{\sqrt{6}}{3}$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(y+\frac{\sqrt{6}}{3}\right)$$

Multiply \frac{1}{3} times y+\frac{\sqrt{6}}{3}.

$$x=\frac{1}{3}y+\frac{\sqrt{6}}{9}$$

Substitute \frac{y}{3}+\frac{\sqrt{6}}{9} for x in the other equation, x+3y=\sqrt{9}.

$$\frac{1}{3}y+\frac{\sqrt{6}}{9}+3y=\sqrt{9}$$

Add \frac{y}{3} to 3y.

$$\frac{10}{3}y+\frac{\sqrt{6}}{9}=\sqrt{9}$$

Subtract \frac{\sqrt{6}}{9} from both sides of the equation.

$$\frac{10}{3}y=-\frac{\sqrt{6}}{9}+3$$

Divide both sides of the equation by \frac{10}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{\sqrt{6}}{30}+\frac{9}{10}$$

Substitute \frac{9}{10}-\frac{\sqrt{6}}{30} for y in x=\frac{1}{3}y+\frac{\sqrt{6}}{9}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{3}\left(-\frac{\sqrt{6}}{30}+\frac{9}{10}\right)+\frac{\sqrt{6}}{9}$$

Multiply \frac{1}{3} times \frac{9}{10}-\frac{\sqrt{6}}{30}.

$$x=-\frac{\sqrt{6}}{90}+\frac{3}{10}+\frac{\sqrt{6}}{9}$$

Add \frac{\sqrt{6}}{9} to -\frac{\sqrt{6}}{90}+\frac{3}{10}.

$$x=\frac{\sqrt{6}+3}{10}$$

The system is now solved.

$$x=\frac{\sqrt{6}+3}{10},y=-\frac{\sqrt{6}}{30}+\frac{9}{10}$$

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.

$$3x-y=\sqrt{\frac{2}{3}},x+3y=\sqrt{9}$$

To make 3x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 3.

$$3x-y=\sqrt{\frac{2}{3}},3x+3\times 3y=3\sqrt{9}$$

Simplify.

$$3x-y=\frac{\sqrt{6}}{3},3x+9y=9$$

Subtract 3x+9y=9 from 3x-y=\frac{\sqrt{6}}{3} by subtracting like terms on each side of the equal sign.

$$3x-3x-y-9y=\frac{\sqrt{6}}{3}-9$$

Add 3x to -3x. Terms 3x and -3x cancel out, leaving an equation with only one variable that can be solved.

$$-y-9y=\frac{\sqrt{6}}{3}-9$$

Add -y to -9y.

$$-10y=\frac{\sqrt{6}}{3}-9$$

Divide both sides by -10.

$$y=-\frac{\sqrt{6}}{30}+\frac{9}{10}$$

Substitute -\frac{\sqrt{6}}{30}+\frac{9}{10} for y in x+3y=\sqrt{9}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+3\left(-\frac{\sqrt{6}}{30}+\frac{9}{10}\right)=\sqrt{9}$$

Multiply 3 times -\frac{\sqrt{6}}{30}+\frac{9}{10}.

$$x+\frac{27-\sqrt{6}}{10}=\sqrt{9}$$

Subtract \frac{-\sqrt{6}+27}{10} from both sides of the equation.

$$x=\frac{\sqrt{6}+3}{10}$$

The system is now solved.

$$x=\frac{\sqrt{6}+3}{10},y=-\frac{\sqrt{6}}{30}+\frac{9}{10}$$