**Rational Equations: **

*The Recipe of Solving Rational Equations*

- First find the LCD (Least Common Denominator), i.e. the Least Common Multiple of all denominators.
- Multiply your rational equation by the LCD.
- Solve the resulting equation (usually a linear equation or a quadratic equation).
- Plug solutions you obtained from STEP 3 into the original rational equation for $x$ to see if they all satisfy.

*Example*. Solve

$$\frac{1}{3x+6}-\frac{1}{x^2-4}=\frac{3}{x-2}.$$

*Solution*. STEP 1. $3x+6=3(x+2)$ and $x^2-4=(x+2)(x-2)$. Thus the LCD is $3(x+2)(x-2)$.

STEP 2. Multiply the equation the by LCD.

$$\frac{1}{3x+6}\cdot 3(x+2)(x-2)-\frac{1}{x^2-4}\cdot 3(x+2)(x-2)=\frac{3}{x-2}\cdot 3(x+2)(x-2)$$

which results the linear equation.

$$(x-2)-3=3\cdot 3(x+2).$$

The solution of this linear equation is $x=-\frac{23}{8}$.

**Radical Equations:**

*The Recipe of Solving Radical Equations*

- First isolate one radical term in one side.
- Square both sides of the equation.
- If all radical are gone, solve the resulting equation (usually linear or quadratic). If not (in case the radical equation had two radical terms), repeat the steps 1 and 2.
- Plug solutions you obtained from STEP 3 into the original radical equation for $x$ to see if they all satisfy.

*Example*. Solve $5+\sqrt{x+7}=x$.

*Solution*. STEP 1. Isolate the radical term $\sqrt{x+7}$ in the LHS.

$$\sqrt{x+7}=x-5.$$

STEP 2. Square both sides of the resulting equation.

$$x+7=(x-5)^2.$$

This is simplified to the quadratic equation

$$x^2-11x+18=0$$ which is factored to

$$(x-9)(x-2)=0.$$

Hence we obtain the two solutions $x=2,9$. However, not all these solutions may satisfy the original radical equation.

STEP 3. If $x=9$, then $$\mbox{LHS}=5+\sqrt{9+7}=9=\mbox{RHS}.$$ However, if $x=2$ then

$$\mbox{LHS}=5+\sqrt{x+7}=8\ne 2=\mbox{RHS}.$$ Therefore $x=9$ is the only solution to the radical equation.

*Example*. Solve $\sqrt{x-3}+\sqrt{x+5}=4$.

*Solution*. STEP 1. Isolate on radical term in one side, say isolate $\sqrt{x-3}$ in the LHS.

$$\sqrt{x-3}=4-\sqrt{x+5}.$$

STEP 2. Square both side of the resulting equation.

\begin{align*}

x-3&=(4-\sqrt{x+5})^2\\

&=16-8\sqrt{x+5}+(x+5)\\

&=x+21-8\sqrt{x+5}.

\end{align*}

Repeat STEP 1. Isolate the radical term $\sqrt{x+5}$ in the RHS.

$$3=\sqrt{x+5}.$$

Repeat STEP 2. Square both sides of the resulting equation.

$$9=x+5.$$

Hence we obtain the solution $x=4$.

One can readily check that $x=4$ satisfies the original radical equation.