ibanez1608
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- Jul 12, 2013
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(X+1)2+(y-1)2=5
(X-1)2+(Y-2)2=10
What is the solution x and y
(X-1)2+(Y-2)2=10
What is the solution x and y
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- Thread starter ibanez1608
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Hello, ibanez1608!
\(\displaystyle \begin{array}{cccccccc}\text{Simplify [1]:} & x^2+2x+1 + y^2-2y+1 \;=\; 5 & \Rightarrow & x^2 + 2x + y^2 - 2y \;=\; 3 & [3] \\ \text{Simplify [2]:} & x^2 - 2x + 1 + y^2 - 4y + 4 \;=\; 10 & \Rightarrow & x^2-2x+y^2-4y \;=\; 5 & [4] \end{array}\)
\(\displaystyle \text{Subtract [3] - [4]: }\:4x + 2y \:=\:-2 \quad\Rightarrow\quad y \:=\:\text{-}2x-1\;\;[5]\)
\(\displaystyle \text{Substitute into [3]: }\: x^2+2x + (\text{-}2x-1)^2 - 2(\text{-}2x-1) \:=\:3\)
\(\displaystyle \text{Simplify: }\: x^2 + 2x + 4x^2 + 4x + 1 + 4x + 2 \:=\:3\)
. . . . . . . \(\displaystyle 5x^2 + 10x \:=\:0 \quad\Rightarrow\quad 5x(x+2) \:=\:0\)
. . \(\displaystyle \text{Hence: }\: x \;=\;0,\,\text{-}2\)
\(\displaystyle \text{Substitute into [5]: }\:y \;=\;\text{-}1,\:3\)
\(\displaystyle \text{Therefore: }\: (x,\,y) \:=\: (0,\,\text{-}1),\;(\text{-}2,\,3)\)
\(\displaystyle \text{Solve: }\:\begin{array}{ccccc}(x+1)^2 + (y-1)^2 &=& 5 & [1] \\ (x-1)^2 + (y-2)^2 &=& 10 & [2]\end{array}\)
\(\displaystyle \begin{array}{cccccccc}\text{Simplify [1]:} & x^2+2x+1 + y^2-2y+1 \;=\; 5 & \Rightarrow & x^2 + 2x + y^2 - 2y \;=\; 3 & [3] \\ \text{Simplify [2]:} & x^2 - 2x + 1 + y^2 - 4y + 4 \;=\; 10 & \Rightarrow & x^2-2x+y^2-4y \;=\; 5 & [4] \end{array}\)
\(\displaystyle \text{Subtract [3] - [4]: }\:4x + 2y \:=\:-2 \quad\Rightarrow\quad y \:=\:\text{-}2x-1\;\;[5]\)
\(\displaystyle \text{Substitute into [3]: }\: x^2+2x + (\text{-}2x-1)^2 - 2(\text{-}2x-1) \:=\:3\)
\(\displaystyle \text{Simplify: }\: x^2 + 2x + 4x^2 + 4x + 1 + 4x + 2 \:=\:3\)
. . . . . . . \(\displaystyle 5x^2 + 10x \:=\:0 \quad\Rightarrow\quad 5x(x+2) \:=\:0\)
. . \(\displaystyle \text{Hence: }\: x \;=\;0,\,\text{-}2\)
\(\displaystyle \text{Substitute into [5]: }\:y \;=\;\text{-}1,\:3\)
\(\displaystyle \text{Therefore: }\: (x,\,y) \:=\: (0,\,\text{-}1),\;(\text{-}2,\,3)\)
D
Deleted member 4993
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Assuming that "X" was meant to be "x" and "Y" was meant to be "y", notice that you are given equations of two circles - and the intersection points of those two circles would be your solutions.(X+1)2+(y-1)2=5
(X-1)2+(Y-2)2=10
What is the solution x and y
Please share your work with us.
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ibanez1608
New member
- Joined
- Jul 12, 2013
- Messages
- 24
Hello, ibanez1608!
\(\displaystyle \begin{array}{cccccccc}\text{Simplify [1]:} & x^2+2x+1 + y^2-2y+1 \;=\; 5 & \Rightarrow & x^2 + 2x + y^2 - 2y \;=\; 3 & [3] \\ \text{Simplify [2]:} & x^2 - 2x + 1 + y^2 - 4y + 4 \;=\; 10 & \Rightarrow & x^2-2x+y^2-4y \;=\; 5 & [4] \end{array}\)
\(\displaystyle \text{Subtract [3] - [4]: }\:4x + 2y \:=\:-2 \quad\Rightarrow\quad y \:=\:\text{-}2x-1\;\;[5]\)
\(\displaystyle \text{Substitute into [3]: }\: x^2+2x + (\text{-}2x-1)^2 - 2(\text{-}2x-1) \:=\:3\)
\(\displaystyle \text{Simplify: }\: x^2 + 2x + 4x^2 + 4x + 1 + 4x + 2 \:=\:3\)
. . . . . . . \(\displaystyle 5x^2 + 10x \:=\:0 \quad\Rightarrow\quad 5x(x+2) \:=\:0\)
. . \(\displaystyle \text{Hence: }\: x \;=\;0,\,\text{-}2\)
\(\displaystyle \text{Substitute into [5]: }\:y \;=\;\text{-}1,\:3\)
\(\displaystyle \text{Therefore: }\: (x,\,y) \:=\: (0,\,\text{-}1),\;(\text{-}2,\,3)\)
i love u man
No, Ibanez, you don't "love" him! You love being handed something for doing nothing.i love u man