How do you solve the system $x y = 9$ $xy = 9$ , $x^{2} + y^{2} = 82$ $x^2 + y^2 = 82$ ?

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How do you solve the system $x y = 9$ $xy = 9$ , $x^{2} + y^{2} = 82$ $x^2 + y^2 = 82$ ?

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Answer 1 · 2015-11-27T22:27:09

$(- 1, - 9)$ $(-1, -9)$ and $(1, 9)$ $(1,9)$
$(- 9, - 1)$ $(-9, -1)$ and $(9, 1)$ $(9, 1)$

Explanation:

This system can be solve by substitution

Eq 1: $x y = 9$ $xy= 9$

$\Rightarrow$ $=>$ Solve for $x$ $x$

$\frac{x y}{y} = \frac{9}{y}$ $(xy)/y= 9/y$

$\Rightarrow x = \frac{9}{y}$ $=> color(red)(x= 9/y)$ Substitute into Eq.2

Eq 2: $x^{2} + y^{2} = 82$ $x^2 +y^2= 82$

${(\frac{9}{y})}^{2} + y^{2} = 82$ $color(red)((9/y)^2) +y^2 = 82$

$\Rightarrow$ $=>$ $(\frac{9}{y}) (\frac{9}{y}) + y^{2} = 82$ $(9/y)(9/y)+y^2 = 82$

$\Rightarrow \frac{81}{y^{2}} + y^{2} = 82$ $=>81/y^2 +y^2 = 82$

$\Rightarrow y^{2} (\frac{81}{y^{2}} + y^{2} = 82)$ $=>color(blue)(y^2)(81/y^2+y^2=82)$

$\Rightarrow 81 + y^{4} = 82 y^{2}$ $=>81+y^4 =82y^2$

$\Rightarrow y^{4} - 82 y^{2} + 81 = 0$ $=>y^4 -82y^2+81=0$

Factor $(y^{2} - 81) (y^{2} - 1) = 0$ $(y^2-81)(y^2-1)= 0$

Set each factor equal to $0$ $0$
a) $(y^{2} - 81) = 0$ $( y^2-81)= 0$
$y^{2} = 81$ $y^2 = 81$
$y = - 9, 9$ $y= -9 , 9$

b) $y^{2} - 1 = 0$ $y^2 -1 = 0$
$y^{2} = 1$ $y^2 = 1$
$y = - 1, 1$ $y= -1, 1$

When $y = - 1$ $y= -1$
$x = \frac{9}{- 1} = - 9$ $x= 9/(-1)= -9$ ; $(- 9, - 1)$ $(-9, -1)$
When $y = 1$ $y= 1$
$x = \frac{9}{10 = 9}$ $x= 9/(10= 9$ ; $(9, 1)$ $(9, 1)$

When $y = - 9$ $y= -9$
$x = \frac{9}{- 9} = - 1$ $x= 9/(-9) = -1$ ; $(- 1, - 9)$ $(-1, -9)$

When $y = 9$ $y= 9$
$x = \frac{9}{9} = 1$ $x= 9/(9) = 1$ ; $(1, 9)$ $(1, 9)$

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How do you solve the system $x y = 9$ $xy = 9$ , $x^{2} + y^{2} = 82$ $x^2 + y^2 = 82$ ?

1 Answer

Explanation:

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How do you solve the system xy=9xy = 9, x2+y2=82x^2 + y^2 = 82?

1 Answer

Explanation:

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How do you solve the system $x y = 9$ $xy = 9$ , $x^{2} + y^{2} = 82$ $x^2 + y^2 = 82$ ?