Solve 3quad-6div2+(1/3-1/4)*12+(-3)^2

$$3 \quad - 6 \div 2 + ( \frac { 1 } { 3 } - \frac { 1 } { 4 } ) \times 12 + ( - 3 ) ^ { 2 }$$

Evaluate

$$10$$

Solution Steps

Divide 6 by 2 to get 3.

$$3-3+\left(\frac{1}{3}-\frac{1}{4}\right)\times 12+\left(-3\right)^{2}$$

Subtract 3 from 3 to get 0.

$$\left(\frac{1}{3}-\frac{1}{4}\right)\times 12+\left(-3\right)^{2}$$

Least common multiple of 3 and 4 is 12. Convert \frac{1}{3} and \frac{1}{4} to fractions with denominator 12.

$$\left(\frac{4}{12}-\frac{3}{12}\right)\times 12+\left(-3\right)^{2}$$

Since \frac{4}{12} and \frac{3}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4-3}{12}\times 12+\left(-3\right)^{2}$$

Subtract 3 from 4 to get 1.

$$\frac{1}{12}\times 12+\left(-3\right)^{2}$$

Cancel out 12 and 12.

$$1+\left(-3\right)^{2}$$

Calculate -3 to the power of 2 and get 9.

$$1+9$$

Add 1 and 9 to get 10.

$$10$$

Factor

$$2\times 5$$