Solve 7+8left(7+1+3*4right)divleft(3+4*9right)

$$7+8 \left( 7+1+3 \cdot 4 \right) \div \left( 3+4 \cdot 9 \right)$$

Evaluate

$$\frac{433}{39}\approx 11.102564103$$

Solution Steps

Add 7 and 1 to get 8.

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

Multiply 3 and 4 to get 12.

$$7+\frac{8\left(8+12\right)}{3+4\times 9}$$

Add 8 and 12 to get 20.

$$7+\frac{8\times 20}{3+4\times 9}$$

Multiply 8 and 20 to get 160.

$$7+\frac{160}{3+4\times 9}$$

Multiply 4 and 9 to get 36.

$$7+\frac{160}{3+36}$$

Add 3 and 36 to get 39.

$$7+\frac{160}{39}$$

Convert 7 to fraction \frac{273}{39}.

$$\frac{273}{39}+\frac{160}{39}$$

Since \frac{273}{39} and \frac{160}{39} have the same denominator, add them by adding their numerators.

$$\frac{273+160}{39}$$

Add 273 and 160 to get 433.

$$\frac{433}{39}$$

Factor

$$\frac{433}{3 \cdot 13} = 11\frac{4}{39} = 11.102564102564102$$