How do you solve the system $x^2+y^2=5$ and $y=3x+5$ ?

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Answer 1 · 2016-08-28T22:38:34

Solution set is $(-2,-1)$ or $(-1,2)$

As $y=3x+5$ , $x^2+y^2=5$

$hArrx^2+(3x+5)^2=5$ or

$x^2×9x^2+2×3x×5+25=5$ or

$10x^2+30x+20=0$ or dividing by $10$ , we get

$x^2+3x+2=0$ or

$x^2+x+2x×2=0$ or

$x(x+1)+2(x+1)=0$ or

$(x+2)(x+1)=0$ i.e.

$x=-2$ or $x=-1$

Hence $y=3×(-2)+5=-1$ or

$y=3×(-1)+5=2$

Hence solution set is $(-2,-1)$ or $(-1,2)$

How do you solve the system x^2+y^2=5x^2+y^2=5 and y=3x+5y=3x+5?