How do you solve the following system algebraically: $x^{2} + y^{2} = 25$ $x^2+y^2=25$ , $x - 2 y + 5 = 0$ $x-2y+5=0$ ?

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How do you solve the following system algebraically: $x^{2} + y^{2} = 25$ $x^2+y^2=25$ , $x - 2 y + 5 = 0$ $x-2y+5=0$ ?

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Answer 1 · 2018-05-10T04:08:11

When $y = 0$ $y=0$ , $x = - 5$ $x=-5$
When $y = 0$ $y=0$ , $x = 3$ $x=3$

Explanation:

$x^{2} + y^{2} = 25$ $x^2+y^2=25$ --- (1)
$x - 2 y + 5 = 0$ $x-2y+5=0$ --- (2)

From (2),
$x - 2 y + 5 = 0$ $x-2y+5=0$
$x = 2 y - 5$ $x=2y-5$ --- (3)

Sub (3) into (1)
${(2 y - 5)}^{2} + y^{2} = 25$ $(2y-5)^2+y^2=25$
$4 y^{2} - 20 y + 25 + y^{2} = 25$ $4y^2-20y+25+y^2=25$
$5 y^{2} - 20 y = 0$ $5y^2-20y=0$
$y^{2} - 4 y = 0$ $y^2-4y=0$
$y (y - 4) = 0$ $y(y-4)=0$
$y = 0, 4$ $y=0,4$

If $y = 0$ $y=0$ ,
$x^{2} + 0^{2} = 25$ $x^2+0^2=25$
$x = \pm 5$ $x=+-5$
But, if you sub $y = 0$ $y=0$ into (3), only $x = - 5$ $x=-5$ works

If $y = 4$ $y=4$
$x^{2} + 4^{2} = 25$ $x^2+4^2=25$
$x^{2} = 9$ $x^2=9$
$x = \pm 3$ $x=+-3$
But if you sub $y = 4$ $y=4$ into (3), only $x = 3$ $x=3$ works

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How do you solve the following system algebraically: $x^{2} + y^{2} = 25$ $x^2+y^2=25$ , $x - 2 y + 5 = 0$ $x-2y+5=0$ ?

1 Answer

Explanation:

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How do you solve the following system algebraically: x2+y2=25x^2+y^2=25, x−2y+5=0x-2y+5=0?

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Explanation:

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How do you solve the following system algebraically: $x^{2} + y^{2} = 25$ $x^2+y^2=25$ , $x - 2 y + 5 = 0$ $x-2y+5=0$ ?