What is 64\div 8-2[3+[7-3(3+4-2)]] ?

The solution to 64\div 8-2[3+[7-3(3+4-2)]] is 18

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64\div 8-2[3+[7-3(3+4-2)]]

64 \div 8 - 2 [3 + [7 - 3 (3 + 4 - 2)]]

18

Solution steps

64 \div 8 - 2 [3 + [7 - 3 (3 + 4 - 2)]]

Follow the PEMDAS order of operations

Calculate within parentheses

[3 + [7 - 3 (3 + 4 - 2)]] : - 5

3 + [7 - 3 (3 + 4 - 2)]

Calculate within parentheses

[7 - 3 (3 + 4 - 2)] : - 8

7 - 3 (3 + 4 - 2)

Calculate within parentheses

(3 + 4 - 2) : 5

3 + 4 - 2

Add and subtract (left to right)

3 + 4 = 7

= 7 - 2

7 - 2 = 5

= 5

= 7 - 3 \cdot 5

Multiply and divide (left to right)

3 \cdot 5 : 15

3 \cdot 5

3 \cdot 5 = 15

= 15

= 7 - 15

Add and subtract (left to right)

7 - 15 : - 8

7 - 15

7 - 15 = - 8

= - 8

= - 8

= 3 - 8

Add and subtract (left to right)

3 - 8 : - 5

3 - 8

3 - 8 = - 5

= - 5

= - 5

= 64 \div 8 - 2 (- 5)

Multiply and divide (left to right)

64 \div 8 : 8

64 \div 8

64 \div 8 = 8

= 8

= 8 - 2 (- 5)

Multiply and divide (left to right)

2 (- 5) : - 10

2 (- 5)

Apply rule

a \cdot (- b) = - a \cdot b

2 (- 5) = - 2 \cdot 5 = - 10

= - 10

= 8 - (- 10)

Add and subtract (left to right)

8 - (- 10) : 18

8 - (- 10)

Apply rule

- (- a) = + a

- (- 10) = + 10

= 8 + 10

8 + 10 = 18

= 18

= 18

2 0^{5} \div 4^{5}

- \frac{2}{3} 3 - 5

(- 5)^{2} \times 3 + 12

1 - (9 - 4) \div 5

4 \times 5 - (2^{2} - 2)

What is 64\div 8-2[3+[7-3(3+4-2)]] ?
The solution to 64\div 8-2[3+[7-3(3+4-2)]] is 18