What is x_{1}=((sqrt(5)))/3 ,x^2=((sqrt(5)))/5 ?

The solution to x_{1}=((sqrt(5)))/3 ,x^2=((sqrt(5)))/5 is

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x_{1}=((sqrt(5)))/3 ,x^2=((sqrt(5)))/5

x_{1} = \frac{( 5 )}{3}, x^{2} = \frac{( 5 )}{5}

x = - \frac{1}{5}, x = \frac{1}{5}, x_{1} = \frac{5}{3} x_{1} = \frac{5}{3}

Solution steps

x_{1} = \frac{( 5 )}{3} x^{2} = \frac{( 5 )}{5}

Solve

x^{2} = \frac{( 5 )}{5} : x = - \frac{1}{5}, x = \frac{1}{5}

For

x_{1} = \frac{( 5 )}{3}

, subsitute

x

with

- \frac{1}{5} : x_{1} = \frac{5}{3}

For

x_{1} = \frac{( 5 )}{3}

, subsitute

x

with

\frac{1}{5} : x_{1} = \frac{5}{3}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x_{1} = \frac{( 5 )}{3}, x^{2} = \frac{( 5 )}{5}

are

x = - \frac{1}{5}, x = \frac{1}{5}, x_{1} = \frac{5}{3} x_{1} = \frac{5}{3}

2 x - y = 5, - 2 x + y = - 5

x = 7 - 3 y, x^{2} - y^{2} = 39

2 x - 3 y = 7, (a + b) x - (a + b - 3) y = 4 a + b

3 a + 2 = 3, b + 4 = 3 c + 5

\frac{1}{x} + \frac{4}{y} = \frac{2}{7}, \frac{4}{3} x + \frac{1}{2} y = - \frac{8}{5}

What is x_{1}=((sqrt(5)))/3 ,x^2=((sqrt(5)))/5 ?
The solution to x_{1}=((sqrt(5)))/3 ,x^2=((sqrt(5)))/5 is