What is x+y=8,xy= 25/9 ?

The solution to x+y=8,xy= 25/9 is

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x+y=8,xy= 25/9

x + y = 8, x y = \frac{25}{9}

(x = 8 - \frac{12 - 119}{3}, x = 8 - \frac{12 + 119}{3}, y = \frac{12 - 119}{3} y = \frac{12 + 119}{3})

Solution steps

[x + y = 8 x y = \frac{25}{9}]

Isolate

x

for

x + y = 8 : x = 8 - y

Plug the solutions

x = 8 - y

into

x y = \frac{25}{9}

For

x y = \frac{25}{9}

, subsitute

x

with

8 - y : y = \frac{12 - 119}{3}, y = \frac{12 + 119}{3}

Plug the solutions

y = \frac{12 - 119}{3}, y = \frac{12 + 119}{3}

into

x + y = 8

For

x + y = 8

, subsitute

y

with

\frac{12 - 119}{3} : x = 8 - \frac{12 - 119}{3}

For

x + y = 8

, subsitute

y

with

\frac{12 + 119}{3} : x = 8 - \frac{12 + 119}{3}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x + y = 8, x y = \frac{25}{9}

are

(x = 8 - \frac{12 - 119}{3}, x = 8 - \frac{12 + 119}{3}, y = \frac{12 - 119}{3} y = \frac{12 + 119}{3})

p = - 20, 0 + \frac{1}{4} q^{2} p = 1000 - \frac{1}{2} q^{2}

2^{x} + 2^{y} = 40, 2^{x + y} = 256

a = \frac{2}{3} b, c = 12

4 x^{2} - 9 (y + 1)^{2} = 36, 2 x + 3 y = - 5

x^{2} - 25 = 0, y^{3} = 216