x^2+y^2=2 y+3x=-2
plz help.. two problems due in 25 min
non linear equations
what are you trying to do? solve for y?
no im trying to solve the systems of equations
points x,y and there may be more than one
Ive got this and the exact same type of problem then im done but there due in 20 min and i ahve no clue how to do it
Hey myninya can you talk on group chat after u answer this thx :)
y+3x=-2 implies y=-2-3x the other the equation we have x^2+y^2=2 lets plug that first equation y=-2-3x into the second equation x^2+y^2=2 x^2+(-2-3x)^2=2 x^2+(4+2(-2)(-3x)+9x^2)=2 x^2+4+12x+9x^2=2 10x^2+12x+4-2=0 10x^2+12x+2=0 5x^2+6x+1=0 you can solve for x right?
okay, they intersect at (-1.7,3.1)
(17/10,and 31/10) is that right??
5x^2+5x+x+1=0 5x(x+1)+(x+1)=0 (5x+1)(x+1)=0 5x=-1 x=-1 x=-1/5 they intersect at two points we can find the corresponding values
(-1/5, -2-3(-1/5))=(-1/5,-2+3/5)=(-1/5,-7/5) (-1,-2-3(-1))=(-1,-2+3)=(-1,1)
I hate to be a pain in the you know what but would you mind helping me with one more before you go, this is a huge help!!!
x^2+y^2=30 x-y^2=-10
okay you do exactly the same steps are before but this time y = root(x+10), okay?
so u get x² + x + 10 = 30, x² + x - 20 = 0, then (x+5)(x-4) = 0, x = -5, x = 4
and y would be root (x+10)
so you plug that into it how??
2nd equation we have y^2=10+x 1st equation we have x^2+10+x=30 x^2+x+10-30=0 x^2+x-20=0 (x+5)(x-4)=0 gives x=-5,4 y^2=10+x so y=+ or - sqrt(10+x) y=+sqrt(10+x) y=sqrt(10+-5)=sqrt(5) y=sqrt(10+4)=sqrt(14) also we have y=-sqrt(10+x) y=-sqrt(10-5)=-sqrt(5) y=-sqrt(10+4)=-sqrt(14) so we have 4 points of intersection
(-5,sqrt(10)) (-5,-sqrt(10)) (4,sqrt(10)) (4,-sqrt(10))
myininaya you seem like a very nice person, can we talk on group chat if ur dont answering this ques?
its still saying my answer is wrong??
nm
wait my y^2 was wrong y^2=x-10 is the mistake
so it wuld b 4roto 14 I got it thank you gor you help
no wait it was right it was -10 not 10 nvm i still stand by my answers
ok thanks got it
:) ok cool
can u help me too plz
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