Mathematics

how do you solve this equation?
(x-4)+y2=4
x2-8x+y2=-12


Answers

yassautumn

4 years ago Comment

[latex](x-4)+y^2=4\\ x^2-8x+y^2=-12\\\\ y^2=4-(x-4)=4-x+4=-x+8\\ x^2-8x+y^2=-12\\\\ x^2-8x-x+8=-12\\ x^2-9x+20=0\\ x^2-4x-5x+20=0\\ x(x-4)-5(x-4)=0\\ (x-5)(x-4)=0\\ x=5 \vee x=4\\\\ y^2=-5+8 \vee y^2=-4+8\\ y^2=3 \vee y^2=4\\ y=-\sqrt3 \vee y=\sqrt3 \vee y=-2 \vee y=2\\\\ x=5,y=-\sqrt3\\ x=5,y=\sqrt3\\ x=4,y=-2\\ x=4,y=2 [/latex]

dalmayshuns101

4 years ago Comment

[latex] (x-4) + y^2 = 4 [/latex]
[latex] x^2 - 8x + y^2 = 12 [/latex][latex]x - 4 + y^2 = 4[/latex]
[latex]x + y^2 = 8[/latex][latex]x^2 - 8x + y^2 = -12[/latex]
-          x + [latex]y^2[/latex] = 8
[latex]x^2 - 9x = -20[/latex][latex]x^2 - 9x + 20 = 0[/latex]
[latex](x - 5)(x - 4)[/latex]
x - 5 = 0           x - 4 = 0
x = 5                x = 4
[latex] x + y^2 = 8 [/latex][latex] 5 + y^2 = 8 [/latex]
[latex] y^2 = 3 [/latex]
[latex] \sqrt{y^2} = \sqrt{3} [/latex]
y ≈ +/- 1.73[latex] 4 + y^2 = 8 [/latex]
[latex] y^2 = 4 [/latex]
[latex] \sqrt{y^2} = \sqrt{4} [/latex]
y = +/- 2