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solve this simultaneous equation x^2+y^2=5 2y+x-3

 Jan 15, 2015

Best Answer 

 #2
avatar+129899 
+5

I'm assuming this is supposed to be ..  x^2+y^2=5   2y+x=3  ..if so......

Using the second equation we can rearrange it to x = 3-2y  ... and substituting this into the first equation, we have

(3 - 2y)^2 + y^2 = 5     simplify

4y^2 -12y + 9 + y^2 = 5     subtract 5 from both sides and simplify

5y^2 -12y +4 = 0

(5y -2) (y -2) = 0    and setting each factor to 0, we have that y =2/5 and y =2

And using x = 3-2y  we have that x = 3-2(2/5) = 11/5   and  x = 3-2(2) = -1

So, our solutions are  (11/5, 2/5) and (-1, 2)

BTW......these are the intersection points of a line [2x + y =3] and a circle [x^2 + y^2 = 5] centered at the origin with a radius of √5

 

 Jan 15, 2015
 #1
avatar+65 
0

It is 23.5x+4438x

 Jan 15, 2015
 #2
avatar+129899 
+5
Best Answer

I'm assuming this is supposed to be ..  x^2+y^2=5   2y+x=3  ..if so......

Using the second equation we can rearrange it to x = 3-2y  ... and substituting this into the first equation, we have

(3 - 2y)^2 + y^2 = 5     simplify

4y^2 -12y + 9 + y^2 = 5     subtract 5 from both sides and simplify

5y^2 -12y +4 = 0

(5y -2) (y -2) = 0    and setting each factor to 0, we have that y =2/5 and y =2

And using x = 3-2y  we have that x = 3-2(2/5) = 11/5   and  x = 3-2(2) = -1

So, our solutions are  (11/5, 2/5) and (-1, 2)

BTW......these are the intersection points of a line [2x + y =3] and a circle [x^2 + y^2 = 5] centered at the origin with a radius of √5

 

CPhill Jan 15, 2015

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