Solve 7(-2)/-3-2+3/-3+2>6/3^2+2(3)

$$\frac { 7 ( - 2 ) } { - 3 - 2 } + \frac { 3 } { - 3 } + 2 > \frac { 6 } { 3 ^ { 2 } + 2 ( 3 ) }$$

Verify

Solution Steps

Multiply 7 and -2 to get -14.

$$\frac{-14}{-3-2}+\frac{3}{-3}+2>\frac{6}{3^{2}+2\times 3}$$

Subtract 2 from -3 to get -5.

$$\frac{-14}{-5}+\frac{3}{-3}+2>\frac{6}{3^{2}+2\times 3}$$

Fraction \frac{-14}{-5} can be simplified to \frac{14}{5} by removing the negative sign from both the numerator and the denominator.

$$\frac{14}{5}+\frac{3}{-3}+2>\frac{6}{3^{2}+2\times 3}$$

Divide 3 by -3 to get -1.

$$\frac{14}{5}-1+2>\frac{6}{3^{2}+2\times 3}$$

Convert 1 to fraction \frac{5}{5}.

$$\frac{14}{5}-\frac{5}{5}+2>\frac{6}{3^{2}+2\times 3}$$

Since \frac{14}{5} and \frac{5}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{14-5}{5}+2>\frac{6}{3^{2}+2\times 3}$$

Subtract 5 from 14 to get 9.

$$\frac{9}{5}+2>\frac{6}{3^{2}+2\times 3}$$

Convert 2 to fraction \frac{10}{5}.

$$\frac{9}{5}+\frac{10}{5}>\frac{6}{3^{2}+2\times 3}$$

Since \frac{9}{5} and \frac{10}{5} have the same denominator, add them by adding their numerators.

$$\frac{9+10}{5}>\frac{6}{3^{2}+2\times 3}$$

Add 9 and 10 to get 19.

$$\frac{19}{5}>\frac{6}{3^{2}+2\times 3}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{19}{5}>\frac{6}{9+2\times 3}$$

Multiply 2 and 3 to get 6.

$$\frac{19}{5}>\frac{6}{9+6}$$

Add 9 and 6 to get 15.

$$\frac{19}{5}>\frac{6}{15}$$

Reduce the fraction \frac{6}{15} to lowest terms by extracting and canceling out 3.

$$\frac{19}{5}>\frac{2}{5}$$

Compare \frac{19}{5} and \frac{2}{5}.

$$\text{true}$$