Lesson 63: Symbols of Inclusion

New Concept

Parentheses (), brackets [], and braces {} are symbols of inclusion that group parts of an expression. A symbol of inclusion is a symbol that groups numbers together.

What’s next

This is just the foundation. Next, you’ll tackle worked examples with multiple, nested symbols, including division bars and absolute value signs.

Property

When an expression contains multiple symbols of inclusion, such as parentheses (), brackets [], or braces {}, we simplify within the innermost symbols first.

Examples

$50 - [20 + (10 - 5)] = 50 - [20 + 5] = 50 - 25 = 25$
$100 - 3[2(6 - 2)] = 100 - 3[2(4)] = 100 - 3[8] = 100 - 24 = 76$
$30 - [40 - (10 - 2)] = 30 - [40 - 8] = 30 - 32 = -2$

Explanation

Think of these symbols as nested secret boxes. You must always unlock and solve the calculation in the innermost box first before moving outward. Tackling problems from the inside out is the key to getting the right answer without getting lost in all the numbers and operations.

Property

Absolute value symbols may serve as symbols of inclusion. We find the absolute value as the first step of simplifying within the parentheses.

Examples

$12 - (8 - |4 - 6| + 2) = 12 - (8 - |-2| + 2) = 12 - (8 - 2 + 2) = 4$
$12 + 3|8 - 2| = 12 + 3|6| = 12 + 3(6) = 12 + 18 = 30$
$20 - |10 - 5 \cdot 3| = 20 - |10 - 15| = 20 - |-5| = 20 - 5 = 15$

Explanation

These bars are a priority! Solve the math inside them and take the positive result before you do other operations inside the parentheses. It's a special grouping symbol that demands you handle its contents first, turning any result into a positive value for the main calculation.

Property

A division bar can serve as a symbol of inclusion. We simplify above and below the division bar before we divide.

Examples

$\frac{4 + 5 \times 6 - 7}{10 - (9 - 8)} = \frac{4 + 30 - 7}{10 - 1} = \frac{27}{9} = 3$
$\frac{10 + 9 \cdot 8 - 7}{6 \cdot 5 - 4 \cdot 3 + 2} = \frac{10 + 72 - 7}{30 - 12 + 2} = \frac{75}{20} = \frac{15}{4}$

Explanation

Treat the numerator and the denominator as two separate mini-problems. You need to completely solve the top expression and the bottom expression first. Only when you have a single number on top and a single number on bottom can you perform the final, epic division!