Parentheses in simple Order of Operations: Using parentheses

First, we solve the exercise within the parentheses:

$4\cdot2-3:4=$

We place multiplication and division exercises within parentheses:

$(4\cdot2)-(3:4)=$

We solve the exercises within the parentheses:

$8-\frac{3}{4}=7\frac{1}{4}$

To solve the expression $\frac{1}{3}+(2-1)$, we need to follow the order of operations, often abbreviated as PEMDAS (Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)). This problem primarily involves parentheses and addition.

We'll start by solving the expression within the parentheses:

• The expression inside the parentheses is $(2-1)$. Subtracting, we get:

$2 - 1 = 1$

After solving the parentheses, the expression becomes:

$\frac{1}{3} + 1$

Next, we perform the addition:

• Since $1$ can be written as $\frac{3}{3}$ to have a common denominator with $\frac{1}{3}$, we add:

$\frac{1}{3} + \frac{3}{3} = \frac{1+3}{3} = \frac{4}{3}$

The fraction $\frac{4}{3}$ can also be expressed as a mixed number:

• $\frac{4}{3} = 1 \frac{1}{3}$ (where $1$ is the whole number and $\frac{1}{3}$ is the fractional part)

Thus, the correct answer is $1\frac{1}{3}$.

We solve the exercise in parentheses:

$3:9\cdot9-6=$

$\frac{3}{9}\cdot9-6=$

We simplify and subtract:

$3-6=-3$

We solve the exercise in parentheses:$3\cdot3+5:1=$

We place in parentheses the multiplication and division exercises:

$(3\cdot3)+(5:1)=$

We solve the exercises in parentheses:

$9+5=14$