Let's define a few terms.
Integer: The whole numbers and their opposites
Whole numbers: The set of all positive numbers and 0.
Rational Number: A number in which its decimal expansion either terminates or repeats. Some people also know that a rational number can always be expressed as \(\frac{a}{b}\) such that \(a,b\in\mathbb{Z}\hspace{1mm}\text{and}\hspace{1mm}GCF(a,b)=1\).
Irrational Number: A number in which its decimal expansion neither terminates nor repeats. Some people also know that an irrational number canot be expressed as \(\frac{a}{b}\) such that \(a,b\in\mathbb{Z}\).
Now that we have defined these terms, let's figure out the classification of all these numbers using a table!
\(\sqrt{32}=4\sqrt{2}\) | \(-4.2\) | \(\frac{14}{5}=2.8\) | \(0\) | \(\sqrt{100}=10\) | \(0.\overline{1936}\) | \(1\frac{5}{7}=\frac{12}{7}\) | ||
Integer? | ✘ | ✘ | ✘ | ✔ | ✔ | ✘ | ✘ | |
Whole Number? | ✘ | ✘ | ✘ | ✔ | ✔ | ✘ | ✘ | |
Rational Number? | ✘ | ✔ | ✔ | ✔ | ✔ | ✔ | ✔ | |
Irrational Number? | ✔ | ✘ | ✘ | ✘ | ✘ | ✘ | ✘ | |
I will solve for x in the equation \(\frac{1}{8}(56x-24)=\frac{8}{7}(21x-7)+7\).
\(\frac{1}{8}(56x-24)=\frac{8}{7}(21x-7)+7\) | Distribute the 1/8 to both terms inside of the parentheses. |
\(7x-3=\frac{8}{7}(21x-7)+7\) | Multiply both sides by 7 to get rid of all the fractions in this equation. |
\(49x-21=8(21x-7)+49\) | Inside the parentheses, let's factor our a GCF of 7 from both terms of the parentheses. |
\(8(21x-7)=(8*7)(3x-1)=56(3x-1)\) | Now, plug that back into the equation. |
\(49x-21=56(3x-1)+49\) | Divide all sides of the equation by its GCF, 7. This should ease computation since the numbers will be easier to work with. |
\(7x-3=8(3x-1)+7\) | Distribute the 8 to both terms in the parentheses. |
\(7x-3=24x-8+7\) | Simplify the left hand side. |
\(7x-3=24x-1\) | Subtract 7x on both sides. |
\(-3=17x-1\) | Add 1 to both sides. |
\(-2=17x\) | Divide by 17 on both sides. |
\(x=-\frac{2}{17}\) |