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

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Answer 1 · 2018-05-10T04:08:11

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

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

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

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

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

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

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