A variable is a placeholder for an unknown. They can be English or Greek letters and symbols. In the equation below, you can identify the following:
\(y = 5x + 3 \)
\(\boldsymbol{y}\) and \(\boldsymbol{x}\) are the variables.
\(\boldsymbol{5}\) before the \(x\) is known as the coefficient. After combining the two, \(5x\), is referred to as a term.
\(\boldsymbol{'+'}\) sign is the operator.
\(\boldsymbol{3}\) is a constant and is also a term within the equation.
If you know the value the variable is equal to, you can substitute in that value for the variable. Substituting means you replace the variable with value. After substituting, you can solve the equation.
Substitute \(x = 2\) for \(y = 5x + 3\).
While following the BEDMAS rule, we just have to plug the numbers into the expression where their corresponding variables are. Every time you see an \(x\), substitute it with \(2\).
\(y = 5\textcolor{red}{x} + 3 \)
\(y = 5(\textcolor{red}{2}) + 3 \)
\(y = 10 + 3 \)
\(y = 13\)
Therefore, we can determine that \(\boldsymbol{y=13}\) when \(x=3\).
First, we can substitute \(-1\) for \(x\):
\(y = 3\textcolor{red}{x} + 5 - \textcolor{red}{x}\)
\(y = 3(\textcolor{red}{-1}) + 5 - (\textcolor{red}{-1})\)
Next, we can use BEDMAS to evaluate the expression:
\(y = -3 + 5 + 1\)
\(y = 3\)
Therefore, we can determine that \(\boldsymbol{y = 3}\) when \(x = -1\).
Recall that exponential notations consist of a base number powered by an exponent number. Here they can also be expressed with variables \(x^y \) where \(x\) is the base and \(y\) is the exponent.
Exponential notation is a repeated multiplcation. The base is multiplied by itself according to the exponent:
\(5^3 = (5)(5)(5) = 125\)
Here are some rules that you should remember!
| Name | Law |
| Product | \(x^y \cdot x^z = x^{y+z} \) |
| Power of a Product | \(x^y \cdot z^y = (x \cdot z)^y \) |
| Quotient | \( \cfrac{x^y}{x^z} = x^{y-z}\) |
| Power of a Quotient | \( \cfrac{x^y}{z^y} = \left(\cfrac {x}{z}\right)^y \) |
| Power of a Power | \(x^{y^z} = x^{(y \cdot z)} \) |
| Fractional Exponent | \(\sqrt[z]{{x^y}} = x^{y/z} \) |
| Negative Exponent | \(x^{-y} = \cfrac {1}{x^y} \) |
| Zero Exponent | \(x^0 = 1 \) |
| Zero Base | \(0^y = 0 \); for \(y > 0\) |
Simplify \(x^2 \cdot x^4 \) using exponent rules.
Since the bases of the terms are the same and they are being multiplied, we can add the exponents:
\(= x^{\textcolor{red}{2}+\textcolor{red}{4}}\)
\(= x^6\)
Therefore, we can determine that \(x^2 \cdot x^4 \) simplified using exponent rules is \(\boldsymbol{x^6}\).
\(\cfrac{x^7}{x^3}\)
Since the terms contain the same base and are being divided, we can simplify the expression by subtracting the exponents:
\(= x^{\textcolor{red}{7}-\textcolor{red}{3}}\)
\(= x^4\)
Therefore, we can determine that \(\cfrac{x^7}{x^3}\) simplified using exponent rules is \(\boldsymbol{x^4}\).
\((x^3)^4\)
Since this expression represents a power of a power, we can simplify it multiplying the exponents:
\(= x^{\textcolor{red}{3} \cdot \textcolor{red}{4}}\)
\(= x^{12}\)
Therefore, we can determine that \((x^3)^4\) simplified using exponent rules is \(\boldsymbol{x^{12}}\)