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How do you solve $x + y = 3$ $x+y=3$ and $x = 4 - {(y - 1)}^{2}$ $x = 4-(y-1)^2$ ?

1 Answer

Apr 9, 2016

$(x, y) = (0, 3)$ $(x,y)=(0,3)$
or
$(x, y) = (3, 0)$ $(x,y)=(3,0)$

Explanation:

If $x + y = 3$ $x+y=3$
then $y = 3 - x$ $y=3-x$

and we can substitute $(3 - x)$ $(color(red)(3-x))$ in place of $y$ $y$ in the second equation: $x = 4 - {(y - 1)}^{2}$ $x=4-(y-1)^2$

This gives
$XXX x = 4 - {(3 - x - 1)}^{2}$ $color(white)("XXX")x=4-(color(red)(3-x)-1)^2$

$XXX x = 4 - {(2 - x)}^{2}$ $color(white)("XXX")x=4-(2-x)^2$

$XXX x = 4 - (4 - 4 x + x^{2})$ $color(white)("XXX")x=4-(4-4x+x^2)$

#color(white)("XXX")x= 4x-x^2

$XXX 3 x - x^{2} = 0$ $color(white)("XXX")3x-x^2=0$

$XXX x (3 - x) = 0$ $color(white)("XXX")x(3-x)=0$

$XXX \to x = 0 XX or XX x = 3$ $color(white)("XXX")rarr x=0color(white)("XX")orcolor(white)("XX")x=3$

If $x = 0$ $x=0$ then $x + y = 3 \to y = 3$ $x+y=3 rarr y=3$

If $x = 3$ $x=3$ then $x + y = 3 \to y = 0$ $x+y=3 rarr y=0$

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