How do you solve the system $10 y = x^{2}$ $10y=x^2$ and $x^{2} - 6 = - 2 y$ $x^2-6=-2y$ ?

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Answer 1 · 2016-07-15T07:26:18

We have two solutions $(- \sqrt{5}, \frac{1}{2})$ $(-sqrt5,1/2)$ and $(\sqrt{5}, \frac{1}{2})$ $(sqrt5,1/2)$

As $10 y = x^{2}$ $10y=x^2$ and $x^{2} - 6 = - 2 y$ $x^2-6=-2y$ , putting value of $x^{2}$ $x^2$ from first equation into second we get

$10 y - 6 = - 2 y$ $10y-6=-2y$ or

$10 y + 2 y + 6 - 6 = - 2 y + 2 y + 6$ $10y+2y+6-6=-2y+2y+6$ or

$12 y = 6$ $12y=6$ or

$y = \frac{6}{12} = \frac{1}{2}$ $y=6/12=1/2$ .

Hence, $x^{2} = 10 \times \frac{1}{2} = 5$ $x^2=10×1/2=5$ or

$x^{2} - 5 = 0$ $x^2-5=0$ or

$x^{2} - {(\sqrt{5})}^{2} = 0$ $x^2-(sqrt5)^2=0$ or

$(x - \sqrt{5}) (x + \sqrt{5}) = 0$ $(x-sqrt5)(x+sqrt5)=0$

Hence, either $x - \sqrt{5} = 0$ $x-sqrt5=0$ i.e. $x = \sqrt{5}$ $x=sqrt5$

or $x + \sqrt{5} = 0$ $x+sqrt5=0$ i.e. $x = - \sqrt{5}$ $x=-sqrt5$

Hence we have two solutions $(- \sqrt{5}, \frac{1}{2})$ $(-sqrt5,1/2)$ and $(\sqrt{5}, \frac{1}{2})$ $(sqrt5,1/2)$

How do you solve the system 10y=x^210y=x210y=x^2 and x^2-6=-2yx2−6=−2yx^2-6=-2y?