Solve 3+x(sqrt{3}+1)=x(1-sqrt{2})+5+sqrt{6}

$$3 + x ( \sqrt { 3 } + 1 ) = x ( 1 - \sqrt { 2 } ) + 5 + \sqrt { 6 }$$

Solve for x

$$x=\sqrt{2}\approx 1.414213562$$

Steps for Solving Linear Equation

Use the distributive property to multiply x by \sqrt{3}+1.

$$3+x\sqrt{3}+x=x\left(1-\sqrt{2}\right)+5+\sqrt{6}$$

Use the distributive property to multiply x by 1-\sqrt{2}.

$$3+x\sqrt{3}+x=x-x\sqrt{2}+5+\sqrt{6}$$

Subtract x from both sides.

$$3+x\sqrt{3}+x-x=-x\sqrt{2}+5+\sqrt{6}$$

Combine x and -x to get 0.

$$3+x\sqrt{3}=-x\sqrt{2}+5+\sqrt{6}$$

Add x\sqrt{2} to both sides.

$$3+x\sqrt{3}+x\sqrt{2}=5+\sqrt{6}$$

Subtract 3 from both sides.

$$x\sqrt{3}+x\sqrt{2}=5+\sqrt{6}-3$$

Subtract 3 from 5 to get 2.

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

Combine all terms containing x.

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

The equation is in standard form.

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

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

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

Dividing by \sqrt{3}+\sqrt{2} undoes the multiplication by \sqrt{3}+\sqrt{2}.

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

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

$$x=\sqrt{2}$$