What is [3+(9/3-8)/(7-4+2)] ?

The solution to [3+(9/3-8)/(7-4+2)] is 2

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[3+(9/3-8)/(7-4+2)]

[3 + (9/3 - 8) / (7 - 4 + 2)]

2

Solution steps

3 + (9/3 - 8) / (7 - 4 + 2)

Follow the PEMDAS order of operations

Calculate within parentheses

(9/3 - 8) : - 5

9/3 - 8

Multiply and divide (left to right)

9/3 : 3

9/3

9/3 = 3

= 3

= 3 - 8

Add and subtract (left to right)

3 - 8 : - 5

3 - 8

3 - 8 = - 5

= - 5

= - 5

= 3 - 5/ (7 - 4 + 2)

Calculate within parentheses

(7 - 4 + 2) : 5

7 - 4 + 2

Add and subtract (left to right)

7 - 4 = 3

= 3 + 2

3 + 2 = 5

= 5

= 3 - 5/ (7 - 4 + 2)

Multiply and divide (left to right)

5/ (7 - 4 + 2) : \frac{5}{( 7 - 4 + 2 )}

5/ (7 - 4 + 2)

5/ (7 - 4 + 2) = \frac{5}{( 7 - 4 + 2 )}

= \frac{5}{( 7 - 4 + 2 )}

= 3 - \frac{5}{( 7 - 4 + 2 )}

Add and subtract (left to right)

3 - \frac{5}{( 7 - 4 + 2 )} : 2

3 - \frac{5}{( 7 - 4 + 2 )}

3 - \frac{5}{( 7 - 4 + 2 )} = 2

3 - \frac{5}{7 - 4 + 2}

\frac{5}{7 - 4 + 2} = 1

\frac{5}{7 - 4 + 2}

Add/Subtract the numbers:

7 - 4 + 2 = 5

= \frac{5}{5}

Apply rule

\frac{a}{a} = 1

= 1

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= 2

5^{2} - 2^{3} (- 2)^{2} + 6^{0}

- \frac{2}{3} (2) + 3

(- 1)^{2} - 7

5 \cdot 4 + 3

14/4 - (3/11 + 1/11 + 2111)