Solve sqrt{2x+4}=sqrt{y-3x}

$$\sqrt{ 2x+4 } = \sqrt{ y-3x }$$

Solve for x

$$x=\frac{y-4}{5}$$

$$y\geq -6$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+4}\right)^{2}=\left(\sqrt{y-3x}\right)^{2}$$

Calculate \sqrt{2x+4} to the power of 2 and get 2x+4.

$$2x+4=\left(\sqrt{y-3x}\right)^{2}$$

Calculate \sqrt{y-3x} to the power of 2 and get y-3x.

$$2x+4=y-3x$$

Add 3x to both sides.

$$2x+4+3x=y$$

Combine 2x and 3x to get 5x.

$$5x+4=y$$

Subtract 4 from both sides.

$$5x=y-4$$

Divide both sides by 5.

$$\frac{5x}{5}=\frac{y-4}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x=\frac{y-4}{5}$$

Substitute \frac{y-4}{5} for x in the equation \sqrt{2x+4}=\sqrt{y-3x}.

$$\sqrt{2\times \frac{y-4}{5}+4}=\sqrt{y-3\times \frac{y-4}{5}}$$

Simplify. The value x=\frac{y-4}{5} satisfies the equation.

$$\left(\frac{2}{5}y+\frac{12}{5}\right)^{\frac{1}{2}}=\left(\frac{2}{5}y+\frac{12}{5}\right)^{\frac{1}{2}}$$

Equation \sqrt{2x+4}=\sqrt{y-3x} has a unique solution.

$$x=\frac{y-4}{5}$$

Solve for y

$$y=5x+4$$

$$x\geq -2$$