Solve 3x+sqrt{a+b}=y

$$3 x + \sqrt { a + b } = y$$

Solve for a

$$a=\left(3x-y\right)^{2}-b$$

$$-\left(3x-y\right)\geq 0$$

Steps for Solving Radical Equations

Subtract 3x from both sides of the equation.

$$\sqrt{a+b}+3x-3x=y-3x$$

Subtracting 3x from itself leaves 0.

$$\sqrt{a+b}=y-3x$$

Square both sides of the equation.

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

Subtract b from both sides of the equation.

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

Subtracting b from itself leaves 0.

$$a=\left(y-3x\right)^{2}-b$$

Solve for b

$$b=\left(3x-y\right)^{2}-a$$

$$-\left(3x-y\right)\geq 0$$

Steps for Solving Radical Equations

Subtract 3x from both sides of the equation.

$$\sqrt{b+a}+3x-3x=y-3x$$

Subtracting 3x from itself leaves 0.

$$\sqrt{b+a}=y-3x$$

Square both sides of the equation.

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

Subtract a from both sides of the equation.

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

Subtracting a from itself leaves 0.

$$b=\left(y-3x\right)^{2}-a$$

Solve for a (complex solution)

$$a=\left(3x-y\right)^{2}-b$$

$$x=\frac{y}{3}\text{ or }arg(3x-y)\geq \pi$$

Solve for b (complex solution)

$$b=\left(3x-y\right)^{2}-a$$

$$x=\frac{y}{3}\text{ or }arg(3x-y)\geq \pi$$