An **equation** is a statement where two expressions are equal. For example \[3 + 5 = 8 \qquad \qquad 4x + 7 = 16\]

To preserve equality (be able to write down "$=$" and have it be true) we have seen one method already: **multiplying by one**. For example:
\begin{align}
\dfrac{1}{6} + \dfrac{1}{8} &= 1\cdot \dfrac{1}{6} + \dfrac{1}{8} \cdot 1
\\ &= \dfrac{4}{4}\cdot \dfrac{1}{6} + \dfrac{1}{8}\cdot \dfrac{3}{3}
\end{align}

Useful for introducing something you need, such as LCD or rationalizing!

There are also operations of equality. If you $+, -, \times, \div$ by the same number or expression on each side, equality is also preserved.

**Solving an equation** means to find all values of the variables which makes the equation true.

For the equation $3x + 4 = 10$, what value of $x$ solves the equation?

To solve the above equation, you need to end with the statement \[x = \cdots\]

This process of getting the variable by itself is called **isolating the variable**.

Solve the equation $7x - 4 = 3x + 8$.

The previous equation is called a **linear equation**, which is an equation that looks like \[ax + b = 0\] with perhaps some manipulations.

Solve for (isolate) $M$ in the equation $F = G\dfrac{mM}{r^2}$.

In general, if you want to isolate a variable, let's say "$x$", there are four steps:

- Expand all expressions into terms so there are no parentheses.
- Collect all terms with $x$ on one side. Put all other terms on the other.
- Convert $x$ into a factor by using the GCF factoring method.
- Divide both sides by the factor attached to $x$, therefore isolating $x$.

Isolate $w$ in the equation $A = 2lw + 2wh + 2lh$.

Isolate $x$ in the equation $a - b(c + dx) = ex$.

The previous technique doesn't work if we have the variable to both a second power and a first power like \[x^2 + 5x = 1\] if we try factoring: \[x(x + 5) = 1 \implies x = \dfrac{1}{x + 5}\]

But $x$ is on both sides. Here's how to solve these, called **quadratic equations**:

A **quadratic equation** is an equation of the form \[ax^2 + bx + c = 0\] where $a,b,c \in \mathbb{R}$ and $a \neq 0$.

The following property is one technique to solve quadratics:

The **zero-product property** says \[A\cdot B = 0 \qquad \text{if and only if}\qquad A = 0 \quad \text{or} \quad B = 0\]

Find all real-valued solutions (meaning your solutions $x \in \mathbb{R}$) of the equation $x^2 + 5x = 24$.

Another technique to solve quadratics is by using this idea:

The solutions of the equation \[x^2 = c\] are $x = \sqrt{c}$ and $x = - \sqrt{c}$.

Find all real solutions of both equations:

- $x^2 = 5$
- $(x-4)^2 = 5$

Here is one more method to solve quadratics.

The solutions of the quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$ are \[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Find all real-valued solutions of each equation.

- $3x^2 - 5x - 1 = 0$
- $4x^2 + 12x + 9 = 0$
- $x^2 + 2x + 2 = 0$

There are other types of equations, not just linear or quadratic, such as \[\dfrac{3}{x} - \dfrac{2}{x - 3} = -\dfrac{12}{x^2 - 9} \qquad \qquad \sqrt{4x - 3} = 5 + x \qquad \qquad x^4 - 5x^2 + 4 = 0\]

Oftentimes, what will happen is this:

- You start with a equation that isn't linear or quadratic, like above.
- Try to convert the equation so that it looks linear or quadratic.
- If it's linear, follow the four step process for isolating the variable.
- If it's quadratic, choose one of three quadratic solving methods from above.

For these, think about "rescuing $x$" from the denominator, making it a global term.

Solve the equation $\dfrac{3}{x} - \dfrac{2}{x - 3} = -\dfrac{12}{x^2 - 9}$.

Solve the equation $\dfrac{1}{x - 1} - \dfrac{2}{x^2} = 0$.

For these, think about removing the root to "rescue $x$", making it a global term.

Solve the equation $2x = 1 - \sqrt{2 - x}$.

Solve the equation $\sqrt{3x + 1} = 2 + \sqrt{x + 1}$.