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

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Answer 1 · 2018-06-19T21:38:17

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$x = y + 3 (1)$ $x = y+3" " color(red)((1)$

$x - y^{2} = 3 y (2)$ $x - y^2 = 3y " " color(blue)((2)$

Substitute $(1)$ $color(red)((1)$ into $(2)$ $color(blue)((2)$

$\Rightarrow (y + 3) - y^{2} = 3 y$ $=> (y+3 ) - y^2 = 3y$

$\Rightarrow y^{2} + 2 y - 3 = 0$ $=> y^2 +2y -3 = 0$

$\Rightarrow (y - 1) (y + 3) = 0$ $=> ( y-1)(y+3) = 0$

$\Rightarrow y = 1, y = - 3$ $=> y = 1 , y =-3$

Using $(1)$ $color(red)((1)$

$y = 1 \Rightarrow x = 1 + 3 = 4$ $y=1 => x = 1+3 = 4$

$y = - 3 \Rightarrow x = - 3 + 3 = 0$ $y =-3 => x = -3 + 3 = 0$

Hence the solutions are $x = 4, y = 1$ $x = 4 , y=1$ and $x = 0, y = - 3$ $x = 0 , y=-3$

In coordinate form:

$(4, 1)$ $(4,1)$ and $(0, - 3)$ $(0,-3)$

How do you solve x=y+3x=y+3x=y+3 and x-y^(2)=3yx−y2=3yx-y^(2)=3y using substitution?