Solve {l}{sqrt{2}x+sqrt{3}x=6}{sqrt{3}x+sqrt{8}y=0}

$$\left. \begin{array} { l } { \sqrt { 2 } x + \sqrt { 3 } x = 6 } \\ { \sqrt { 3 } x + \sqrt { 8 } y = 0 } \end{array} \right.$$

Solve for x, y

$$x=6\left(\sqrt{3}-\sqrt{2}\right)\approx 1.907023471$$

$$y=\frac{3\sqrt{2}\left(\sqrt{6}-3\right)}{2}\approx -1.167808608$$

Steps Using Substitution

Consider the first equation. Combine all terms containing x,y.

$$\left(\sqrt{2}+\sqrt{3}\right)x=6$$

Consider the second equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\sqrt{3}x+2\sqrt{2}y=0$$

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.

$$\left(\sqrt{2}+\sqrt{3}\right)x=6,\sqrt{3}x+2\sqrt{2}y=0$$

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.

$$\left(\sqrt{2}+\sqrt{3}\right)x=6$$

Divide both sides by \sqrt{2}+\sqrt{3}.

$$x=6\sqrt{3}-6\sqrt{2}$$

Substitute -6\sqrt{2}+6\sqrt{3} for x in the other equation, \sqrt{3}x+2\sqrt{2}y=0.

$$\sqrt{3}\left(6\sqrt{3}-6\sqrt{2}\right)+2\sqrt{2}y=0$$

Multiply \sqrt{3} times -6\sqrt{2}+6\sqrt{3}.

$$18-6\sqrt{6}+2\sqrt{2}y=0$$

Subtract -6\sqrt{6}+18 from both sides of the equation.

$$2\sqrt{2}y=6\sqrt{6}-18$$

Divide both sides by 2\sqrt{2}.

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

The system is now solved.

$$x=6\sqrt{3}-6\sqrt{2},y=-\frac{9\sqrt{2}}{2}+3\sqrt{3}$$