Solve [7/4+8/3]+10/3=7/4[frac{8}{3}+frac{10}{3}]

$$[ \frac { 7 } { 4 } + \frac { 8 } { 3 } ] + \frac { 10 } { 3 } = \frac { 7 } { 4 } [ \frac { 8 } { 3 } + \frac { 10 } { 3 } ]$$

Verify

Solution Steps

Least common multiple of 4 and 3 is 12. Convert \frac{7}{4} and \frac{8}{3} to fractions with denominator 12.

$$\frac{21}{12}+\frac{32}{12}+\frac{10}{3}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Since \frac{21}{12} and \frac{32}{12} have the same denominator, add them by adding their numerators.

$$\frac{21+32}{12}+\frac{10}{3}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Add 21 and 32 to get 53.

$$\frac{53}{12}+\frac{10}{3}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Least common multiple of 12 and 3 is 12. Convert \frac{53}{12} and \frac{10}{3} to fractions with denominator 12.

$$\frac{53}{12}+\frac{40}{12}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Since \frac{53}{12} and \frac{40}{12} have the same denominator, add them by adding their numerators.

$$\frac{53+40}{12}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Add 53 and 40 to get 93.

$$\frac{93}{12}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Reduce the fraction \frac{93}{12} to lowest terms by extracting and canceling out 3.

$$\frac{31}{4}=\frac{7}{4}\left(\frac{8}{3}+\frac{10}{3}\right)$$

Since \frac{8}{3} and \frac{10}{3} have the same denominator, add them by adding their numerators.

$$\frac{31}{4}=\frac{7}{4}\times \frac{8+10}{3}$$

Add 8 and 10 to get 18.

$$\frac{31}{4}=\frac{7}{4}\times \frac{18}{3}$$

Divide 18 by 3 to get 6.

$$\frac{31}{4}=\frac{7}{4}\times 6$$

Express \frac{7}{4}\times 6 as a single fraction.

$$\frac{31}{4}=\frac{7\times 6}{4}$$

Multiply 7 and 6 to get 42.

$$\frac{31}{4}=\frac{42}{4}$$

Compare \frac{31}{4} and \frac{42}{4}.

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