What is 2x^3y=1,0.93xy=1.1 ?

The solution to 2x^3y=1,0.93xy=1.1 is

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2x^3y=1,0.93xy=1.1

2 x^{3} y = 1, 0.93 x y = 1.1

(x = \frac{93}{2 55}, x = - \frac{93}{2 55}, y = \frac{220 55}{93 93} y = - \frac{220 55}{93 93})

Solution steps

[2 x^{3} y = 1 0.93 x y = 1.1]

Isolate

x

for

0.93 x y = 1.1 : x = \frac{110}{93 y}; y \neq = 0

Plug the solutions

x = \frac{110}{93 y}

into

2 x^{3} y = 1

For

2 x^{3} y = 1

, subsitute

x

with

\frac{110}{93 y} : y = \frac{220 55}{93 93}, y = - \frac{220 55}{93 93}

Plug the solutions

y = \frac{220 55}{93 93}, y = - \frac{220 55}{93 93}

into

0.93 x y = 1.1

For

0.93 x y = 1.1

, subsitute

y

with

\frac{220 55}{93 93} : x = \frac{93}{2 55}

For

0.93 x y = 1.1

, subsitute

y

with

- \frac{220 55}{93 93} : x = - \frac{93}{2 55}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

2 x^{3} y = 1, 0.93 x y = 1.1

are

(x = \frac{93}{2 55}, x = - \frac{93}{2 55}, y = \frac{220 55}{93 93} y = - \frac{220 55}{93 93})

8 x^{4} - 3 y = 17, 7 x - 4 y = 8

x - y = - 6, x^{2} - y = - 4

6 x^{2} + 7 x - 3 = 0, x_{1} = x_{2}

x - y = 2, 2 x^{2} + 9 y^{2} = 23

a + b = 4, a^{2} + b^{2} = 2