Solve 2x(sqrt{3}+sqrt{2})-sqrt{2}(3+2x)=sqrt{3}(x+sqrt{2})

$$2 x ( \sqrt { 3 } + \sqrt { 2 } ) - \sqrt { 2 } ( 3 + 2 x ) = \sqrt { 3 } ( x + \sqrt { 2 } )$$

Solve for x

$$x=\sqrt{2}+\sqrt{6}\approx 3.863703305$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x by \sqrt{3}+\sqrt{2}.

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

Use the distributive property to multiply \sqrt{2} by 3+2x.

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

To find the opposite of 3\sqrt{2}+2\sqrt{2}x, find the opposite of each term.

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

Combine 2x\sqrt{2} and -2\sqrt{2}x to get 0.

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

Use the distributive property to multiply \sqrt{3} by x+\sqrt{2}.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

Subtract \sqrt{3}x from both sides.

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

Combine 2x\sqrt{3} and -\sqrt{3}x to get x\sqrt{3}.

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

Add 3\sqrt{2} to both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}x}{\sqrt{3}}=\frac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}}$$

Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.

$$x=\frac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}}$$

Divide \sqrt{6}+3\sqrt{2} by \sqrt{3}.

$$x=\sqrt{2}+\sqrt{6}$$