Solve {l}{A=(4/9-1/3)*(2*6-3^{2})*1/27}{B=21/2*frac{2}{5}+1-(frac{3}{2})^2+(1-frac{1}{2})^2}{Gamma=(225+25)div50+(18-4frac{3}{4}):frac{1}{4}+20^circ+frac{125}{5^3}}

$$\left. \begin{array} { l } { A = ( \frac { 4 } { 9 } - \frac { 1 } { 3 } ) \cdot ( 2 \cdot 6 - 3 ^ { 2 } ) \cdot \frac { 1 } { 27 } } \\ { B = 2 \frac { 1 } { 2 } \cdot \frac { 2 } { 5 } + 1 - ( \frac { 3 } { 2 } ) ^ { 2 } + ( 1 - \frac { 1 } { 2 } ) ^ { 2 } } \\ { \Gamma = ( 225 + 25 ) \div 50 + ( 18 - 4 \frac { 3 } { 4 } ) : \frac { 1 } { 4 } + 20 ^ { \circ } + \frac { 125 } { 5 ^ { 3 } } } \end{array} \right.$$

Solve for A, B, Γ

$$A=\frac{1}{81}\approx 0.012345679$$

$$B=0$$

$$\Gamma =79$$

Solution Steps

Consider the first equation. Subtract \frac{1}{3} from \frac{4}{9} to get \frac{1}{9}.

$$A=\frac{1}{9}\left(2\times 6-3^{2}\right)\times \frac{1}{27}$$

Multiply 2 and 6 to get 12.

$$A=\frac{1}{9}\left(12-3^{2}\right)\times \frac{1}{27}$$

Calculate 3 to the power of 2 and get 9.

$$A=\frac{1}{9}\left(12-9\right)\times \frac{1}{27}$$

Subtract 9 from 12 to get 3.

$$A=\frac{1}{9}\times 3\times \frac{1}{27}$$

Multiply \frac{1}{9} and 3 to get \frac{1}{3}.

$$A=\frac{1}{3}\times \frac{1}{27}$$

Multiply \frac{1}{3} and \frac{1}{27} to get \frac{1}{81}.

$$A=\frac{1}{81}$$

Consider the second equation. Multiply 2 and 2 to get 4.

$$B=\frac{4+1}{2}\times \frac{2}{5}+1-\left(\frac{3}{2}\right)^{2}+\left(1-\frac{1}{2}\right)^{2}$$

Add 4 and 1 to get 5.

$$B=\frac{5}{2}\times \frac{2}{5}+1-\left(\frac{3}{2}\right)^{2}+\left(1-\frac{1}{2}\right)^{2}$$

Multiply \frac{5}{2} and \frac{2}{5} to get 1.

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

Add 1 and 1 to get 2.

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

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$B=2-\frac{9}{4}+\left(1-\frac{1}{2}\right)^{2}$$

Subtract \frac{9}{4} from 2 to get -\frac{1}{4}.

$$B=-\frac{1}{4}+\left(1-\frac{1}{2}\right)^{2}$$

Subtract \frac{1}{2} from 1 to get \frac{1}{2}.

$$B=-\frac{1}{4}+\left(\frac{1}{2}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$B=-\frac{1}{4}+\frac{1}{4}$$

Add -\frac{1}{4} and \frac{1}{4} to get 0.

$$B=0$$

Consider the third equation. Add 225 and 25 to get 250.

$$\Gamma =\frac{250}{50}+\frac{18-\frac{4\times 4+3}{4}}{\frac{1}{4}}+20+\frac{125}{5^{3}}$$

Divide 250 by 50 to get 5.

$$\Gamma =5+\frac{18-\frac{4\times 4+3}{4}}{\frac{1}{4}}+20+\frac{125}{5^{3}}$$

Multiply 4 and 4 to get 16.

$$\Gamma =5+\frac{18-\frac{16+3}{4}}{\frac{1}{4}}+20+\frac{125}{5^{3}}$$

Add 16 and 3 to get 19.

$$\Gamma =5+\frac{18-\frac{19}{4}}{\frac{1}{4}}+20+\frac{125}{5^{3}}$$

Subtract \frac{19}{4} from 18 to get \frac{53}{4}.

$$\Gamma =5+\frac{\frac{53}{4}}{\frac{1}{4}}+20+\frac{125}{5^{3}}$$

Divide \frac{53}{4} by \frac{1}{4} by multiplying \frac{53}{4} by the reciprocal of \frac{1}{4}.

$$\Gamma =5+\frac{53}{4}\times 4+20+\frac{125}{5^{3}}$$

Multiply \frac{53}{4} and 4 to get 53.

$$\Gamma =5+53+20+\frac{125}{5^{3}}$$

Add 5 and 53 to get 58.

$$\Gamma =58+20+\frac{125}{5^{3}}$$

Add 58 and 20 to get 78.

$$\Gamma =78+\frac{125}{5^{3}}$$

Calculate 5 to the power of 3 and get 125.

$$\Gamma =78+\frac{125}{125}$$

Divide 125 by 125 to get 1.

$$\Gamma =78+1$$

Add 78 and 1 to get 79.

$$\Gamma =79$$

The system is now solved.

$$A=\frac{1}{81} B=0 \Gamma =79$$