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:
\(y = 5x + 3 \)
y and x would be the variables
5 before the x is known as the coefficient. While combining the two, \(5x \), will be referred to as a term.
'+' sign is the operator
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. For the above equation, what would \(y\) be when \(x = 2\)?
While following the BEDMAS rule, we just have to plug the numbers into the expression where their corresponding variables are. Everytime you see an \(x\), replace it with a \(2\).
\(y = 5x + 3 \)
\(y = 5(2) + 3 \)
\(y = 10 + 3 \)
\(y = 13\)
The above results show that \(y=13\) when \(x=3\).
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.
Exponent 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!
| \(x^y \cdot x^z = x^{y+z} \) |
| \(x^y \cdot z^y = (x \cdot z)^y \) |
| \( \cfrac{x^y}{x^z} = x^{y-z}\) |
| \( \cfrac{x^y}{z^y} = (\cfrac {x}{z})^y \) |
| \(x^{y^z} = x^{(y \cdot z)} \) |
| \( \sqrt[z]{{x^y}} = x^{y/z} \) |
| \(x^{-y} = \cfrac {1}{x^y} \) |
| \(x^0 = 1 \) |
| \(0^y = 0 \); for \(y > 0\) |
Let's simplify \(x^2 \cdot x^4 \) using exponent rules. When the base is the same and they are being multiplied, we can add the exponents:
\(x^2 \cdot x^4 = x^{2+4} = x^6\)