Solve A_{3}=(1/sqrt{5}*2^5-10/9*2^3+3*2)-(1/5*1-10/9*1+3)

$$A _ { 3 } = ( \frac { 1 } { \sqrt { 5 } } \cdot 2 ^ { 5 } - \frac { 10 } { 9 } \cdot 2 ^ { 3 } + 3 \cdot 2 ) - ( \frac { 1 } { 5 } \cdot 1 - \frac { 10 } { 9 } \cdot 1 + 3 )$$

Solve for A_3

$$A_{3}=\frac{32\sqrt{5}}{5}-\frac{224}{45}\approx 9.333057278$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{1}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$A_{3}=\frac{\sqrt{5}}{\left(\sqrt{5}\right)^{2}}\times 2^{5}-\frac{10}{9}\times 2^{3}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

The square of \sqrt{5} is 5.

$$A_{3}=\frac{\sqrt{5}}{5}\times 2^{5}-\frac{10}{9}\times 2^{3}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Calculate 2 to the power of 5 and get 32.

$$A_{3}=\frac{\sqrt{5}}{5}\times 32-\frac{10}{9}\times 2^{3}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Express \frac{\sqrt{5}}{5}\times 32 as a single fraction.

$$A_{3}=\frac{\sqrt{5}\times 32}{5}-\frac{10}{9}\times 2^{3}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Calculate 2 to the power of 3 and get 8.

$$A_{3}=\frac{\sqrt{5}\times 32}{5}-\frac{10}{9}\times 8+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Express \frac{10}{9}\times 8 as a single fraction.

$$A_{3}=\frac{\sqrt{5}\times 32}{5}-\frac{10\times 8}{9}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Multiply 10 and 8 to get 80.

$$A_{3}=\frac{\sqrt{5}\times 32}{5}-\frac{80}{9}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 5 and 9 is 45. Multiply \frac{\sqrt{5}\times 32}{5} times \frac{9}{9}. Multiply \frac{80}{9} times \frac{5}{5}.

$$A_{3}=\frac{9\sqrt{5}\times 32}{45}-\frac{80\times 5}{45}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Since \frac{9\sqrt{5}\times 32}{45} and \frac{80\times 5}{45} have the same denominator, subtract them by subtracting their numerators.

$$A_{3}=\frac{9\sqrt{5}\times 32-80\times 5}{45}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Do the multiplications in 9\sqrt{5}\times 32-80\times 5.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+3\times 2-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

Multiply 3 and 2 to get 6.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(\frac{1}{5}\times 1-\frac{10}{9}\times 1+3\right)$$

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

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(\frac{1}{5}-\frac{10}{9}\times 1+3\right)$$

Multiply \frac{10}{9} and 1 to get \frac{10}{9}.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(\frac{1}{5}-\frac{10}{9}+3\right)$$

Least common multiple of 5 and 9 is 45. Convert \frac{1}{5} and \frac{10}{9} to fractions with denominator 45.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(\frac{9}{45}-\frac{50}{45}+3\right)$$

Since \frac{9}{45} and \frac{50}{45} have the same denominator, subtract them by subtracting their numerators.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(\frac{9-50}{45}+3\right)$$

Subtract 50 from 9 to get -41.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(-\frac{41}{45}+3\right)$$

Convert 3 to fraction \frac{135}{45}.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\left(-\frac{41}{45}+\frac{135}{45}\right)$$

Since -\frac{41}{45} and \frac{135}{45} have the same denominator, add them by adding their numerators.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\frac{-41+135}{45}$$

Add -41 and 135 to get 94.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+6-\frac{94}{45}$$

Convert 6 to fraction \frac{270}{45}.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+\frac{270}{45}-\frac{94}{45}$$

Since \frac{270}{45} and \frac{94}{45} have the same denominator, subtract them by subtracting their numerators.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+\frac{270-94}{45}$$

Subtract 94 from 270 to get 176.

$$A_{3}=\frac{288\sqrt{5}-400}{45}+\frac{176}{45}$$

Since \frac{288\sqrt{5}-400}{45} and \frac{176}{45} have the same denominator, add them by adding their numerators.

$$A_{3}=\frac{288\sqrt{5}-400+176}{45}$$

Do the calculations in 288\sqrt{5}-400+176.

$$A_{3}=\frac{288\sqrt{5}-224}{45}$$

Divide each term of 288\sqrt{5}-224 by 45 to get \frac{32}{5}\sqrt{5}-\frac{224}{45}.

$$A_{3}=\frac{32}{5}\sqrt{5}-\frac{224}{45}$$

Assign A_3

$$A_{3}≔\frac{32\sqrt{5}}{5}-\frac{224}{45}$$