What is x+3y=1,3x^2+3y^2=43 ?

The solution to x+3y=1,3x^2+3y^2=43 is

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x+3y=1,3x^2+3y^2=43

x + 3 y = 1, 3 x^{2} + 3 y^{2} = 43

(x = 1 - \frac{9 + 1281}{10}, x = 1 - \frac{9 - 1281}{10}, y = \frac{9 + 1281}{30} y = \frac{9 - 1281}{30})

Solution steps

[x + 3 y = 1 3 x^{2} + 3 y^{2} = 43]

Isolate

x

for

x + 3 y = 1 : x = 1 - 3 y

Plug the solutions

x = 1 - 3 y

into

3 x^{2} + 3 y^{2} = 43

For

3 x^{2} + 3 y^{2} = 43

, subsitute

x

with

1 - 3 y : y = \frac{9 + 1281}{30}, y = \frac{9 - 1281}{30}

Plug the solutions

y = \frac{9 + 1281}{30}, y = \frac{9 - 1281}{30}

into

x + 3 y = 1

For

x + 3 y = 1

, subsitute

y

with

\frac{9 + 1281}{30} : x = 1 - \frac{9 + 1281}{10}

For

x + 3 y = 1

, subsitute

y

with

\frac{9 - 1281}{30} : x = 1 - \frac{9 - 1281}{10}

Verify solutions by plugging them into the original equations

Therefore, the final solutions for

x + 3 y = 1, 3 x^{2} + 3 y^{2} = 43

are

(x = 1 - \frac{9 + 1281}{10}, x = 1 - \frac{9 - 1281}{10}, y = \frac{9 + 1281}{30} y = \frac{9 - 1281}{30})

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