Solve 3(y-sqrt{2})<2y+sqrt{2}=

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

Solve for y

$$y<4\sqrt{2}$$

Solution Steps

Use the distributive property to multiply 3 by y-\sqrt{2}.

$$3y-3\sqrt{2}<2y+\sqrt{2}$$

Subtract 2y from both sides.

$$3y-3\sqrt{2}-2y<\sqrt{2}$$

Combine 3y and -2y to get y.

$$y-3\sqrt{2}<\sqrt{2}$$

Add 3\sqrt{2} to both sides.

$$y<\sqrt{2}+3\sqrt{2}$$

Combine \sqrt{2} and 3\sqrt{2} to get 4\sqrt{2}.

$$y<4\sqrt{2}$$