How do you solve the system $x^{2} - 2 y = 1$ $x^2 - 2y = 1$ , $x^{2} + 5 y = 29$ $x^2 + 5y = 29$ ?

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How do you solve the system $x^{2} - 2 y = 1$ $x^2 - 2y = 1$ , $x^{2} + 5 y = 29$ $x^2 + 5y = 29$ ?

1 Answer

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Answer 1 · 2015-12-18T15:37:08

$| x | = 3$ $absx = 3$
$y = 4$ $y = 4$

Explanation:

You can substract the 1st line to the 2nd one, which will make $x^{2}$ $x^2$ disappear. So the 2nd line is now $7 y = 28$ $7y = 28$ and you now know that $y = 4$ $y = 4$ .

You replace $y$ $y$ by its value in the 1st line of the system :

$x^{2} - 2 y = 1 \Leftrightarrow x^{2} - 8 = 1 \Leftrightarrow x^{2} = 9 \Leftrightarrow | x | = 3$ $x^2 - 2y = 1 iff x^2 - 8 = 1 iff x^2 = 9 iff abs(x) = 3$

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How do you solve the system $x^{2} - 2 y = 1$ $x^2 - 2y = 1$ , $x^{2} + 5 y = 29$ $x^2 + 5y = 29$ ?

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How do you solve the system x2−2y=1x^2 - 2y = 1, x2+5y=29x^2 + 5y = 29?

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How do you solve the system $x^{2} - 2 y = 1$ $x^2 - 2y = 1$ , $x^{2} + 5 y = 29$ $x^2 + 5y = 29$ ?