1.5: Equations

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

Solving Equations

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 for a Variable

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:

  1. Expand all expressions into terms so there are no parentheses.
  2. Collect all terms with $x$ on one side. Put all other terms on the other.
  3. Convert $x$ into a factor by using the GCF factoring method.
  4. 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 $a - b(c + dx) = ex$.

Quadratics

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 $a \neq 0$.

Solving Quadratics with Factoring

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 $x^2 + 5x = 24$.

Solving Quadratics of the form $x^2 = c$

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:

The Quadratic Formula

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.

Other Types of Equations

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:

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

Equations Involving Fractional Expressions

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

Equations Involving Radicals

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