1.5: Equations
An equation is a statement where two mathematical expressions are equal.
For example \[3 + 5 = 8 \qquad \qquad 4x + 7 = 16\] are equations.
Solving Equations
Solving an equation means finding all values of the variables which makes the equation true.
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 add or multiply by the same number or expression on each side, equality is also preserved.
Properties of Equality
- Given $A = B$, we also know $A + C = B + C$.
- Given $A = B$, we also know $A\cdot C = B\cdot C$ provided $C \neq 0$.
For the equation $3x + 4 = 10$, find a solution and a non-solution.
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:
A linear equation is an equation that can be manipulated into the form \[ax + b = 0\]
with $a, b \in \mathbb{R}$ and $a \neq 0$.
Solving for (isolating) a Variable
Solve for (isolate) $M$ in the equation $F = G\dfrac{mM}{r^2}$.
In general, if you want to isolate a variable, let's call "$x$", use these four steps:
- Expand all expressions into global terms.
- Collect terms with $x$ on one side. Put all other terms on the other side.
- Convert $x$ into a global factor using the GCF factoring method.
- Divide both sides by the factor attached to $x$, therefore isolating $x$.
Isolate $w$ in the equation $A = 2lw + 2wh + 2lh$.
The next example appears in Calculus.
Isolate $y'$ in the equation $3(y + xy') - 3x^2 + 2y^2 y' = y'$.
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 $a \neq 0$.
We discuss three ways to solve quadratics.
Method 1: 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 $x^2 + 5x = 24$.
Method 2: 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:
Method 3: 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.
- $3x^2 - 5x - 1 = 0$
- $4x^2 + 12x + 9 = 0$
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:
- 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.
Equations Involving Fractional Expressions
For these, think about "rescuing $x$" from the denominator.
Multiply both sides by the LCD to get $x$ out of the denominator.
Solve the equation $\dfrac{1}{x-\dfrac{1}{2}}-\dfrac{2}{x^{2}} = 0$.
The next example appears in Calculus.
Solve the equation $\dfrac{1 - \dfrac{1}{x^2}}{x} = 0$.
Equations Involving Radicals
First, isolate the square root expression on one side.
Then square both sides, and solve normally.
Solve the equation $\dfrac{\sqrt{x} + 1}{x + 1} = 1$
Solve the equation $2x = 1 - \sqrt{2 - x}$.