Solve 9(7+6)+3(3-2)/7(8-4)-4^2

$$\frac { 9 ( 7 + 6 ) + 3 ( 3 - 2 ) } { 7 ( 8 - 4 ) - 4 ^ { 2 } }$$

Evaluate

$$10$$

Solution Steps

Add 7 and 6 to get 13.

$$\frac{9\times 13+3\left(3-2\right)}{7\left(8-4\right)-4^{2}}$$

Multiply 9 and 13 to get 117.

$$\frac{117+3\left(3-2\right)}{7\left(8-4\right)-4^{2}}$$

Subtract 2 from 3 to get 1.

$$\frac{117+3\times 1}{7\left(8-4\right)-4^{2}}$$

Multiply 3 and 1 to get 3.

$$\frac{117+3}{7\left(8-4\right)-4^{2}}$$

Add 117 and 3 to get 120.

$$\frac{120}{7\left(8-4\right)-4^{2}}$$

Subtract 4 from 8 to get 4.

$$\frac{120}{7\times 4-4^{2}}$$

Multiply 7 and 4 to get 28.

$$\frac{120}{28-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{120}{28-16}$$

Subtract 16 from 28 to get 12.

$$\frac{120}{12}$$

Divide 120 by 12 to get 10.

$$10$$

Factor

$$2\times 5$$