Solve 121/5-(35/6+21/2)+(43/10-1frac{2}{3})=.1

$$12 \frac { 1 } { 5 } - ( 3 \frac { 5 } { 6 } + 2 \frac { 1 } { 2 } ) + ( 4 \frac { 3 } { 10 } - 1 \frac { 2 } { 3 } ) = .1$$

Verify

Solution Steps

Multiply both sides of the equation by 30, the least common multiple of 5,6,2,10,3.

$$30\left(\frac{12\times 5+1}{5}-\left(\frac{3\times 6+5}{6}+\frac{2\times 2+1}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Multiply 12 and 5 to get 60.

$$30\left(\frac{60+1}{5}-\left(\frac{3\times 6+5}{6}+\frac{2\times 2+1}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Add 60 and 1 to get 61.

$$30\left(\frac{61}{5}-\left(\frac{3\times 6+5}{6}+\frac{2\times 2+1}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Multiply 3 and 6 to get 18.

$$30\left(\frac{61}{5}-\left(\frac{18+5}{6}+\frac{2\times 2+1}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Add 18 and 5 to get 23.

$$30\left(\frac{61}{5}-\left(\frac{23}{6}+\frac{2\times 2+1}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Multiply 2 and 2 to get 4.

$$30\left(\frac{61}{5}-\left(\frac{23}{6}+\frac{4+1}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Add 4 and 1 to get 5.

$$30\left(\frac{61}{5}-\left(\frac{23}{6}+\frac{5}{2}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Least common multiple of 6 and 2 is 6. Convert \frac{23}{6} and \frac{5}{2} to fractions with denominator 6.

$$30\left(\frac{61}{5}-\left(\frac{23}{6}+\frac{15}{6}\right)\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Since \frac{23}{6} and \frac{15}{6} have the same denominator, add them by adding their numerators.

$$30\left(\frac{61}{5}-\frac{23+15}{6}\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Add 23 and 15 to get 38.

$$30\left(\frac{61}{5}-\frac{38}{6}\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Reduce the fraction \frac{38}{6} to lowest terms by extracting and canceling out 2.

$$30\left(\frac{61}{5}-\frac{19}{3}\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Least common multiple of 5 and 3 is 15. Convert \frac{61}{5} and \frac{19}{3} to fractions with denominator 15.

$$30\left(\frac{183}{15}-\frac{95}{15}\right)+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Since \frac{183}{15} and \frac{95}{15} have the same denominator, subtract them by subtracting their numerators.

$$30\times \frac{183-95}{15}+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Subtract 95 from 183 to get 88.

$$30\times \frac{88}{15}+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Express 30\times \frac{88}{15} as a single fraction.

$$\frac{30\times 88}{15}+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Multiply 30 and 88 to get 2640.

$$\frac{2640}{15}+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Divide 2640 by 15 to get 176.

$$176+3\left(4\times 10+3\right)-10\left(3+2\right)=3$$

Multiply 4 and 10 to get 40.

$$176+3\left(40+3\right)-10\left(3+2\right)=3$$

Add 40 and 3 to get 43.

$$176+3\times 43-10\left(3+2\right)=3$$

Multiply 3 and 43 to get 129.

$$176+129-10\left(3+2\right)=3$$

Add 176 and 129 to get 305.

$$305-10\left(3+2\right)=3$$

Add 3 and 2 to get 5.

$$305-10\times 5=3$$

Multiply -10 and 5 to get -50.

$$305-50=3$$

Subtract 50 from 305 to get 255.

$$255=3$$

Compare 255 and 3.

$$\text{false}$$