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Simplify this algebraic equation
- Thread starter MrJoe2000
- Start date
Hi, can someone tell me how:
(y^2) - 3y = 4x - (x^2)/2 -10
becomes:
2(x - 4) + (2y - 3)^2 = 1
if it does at all?
MrJoe2000,
one of the reasons that the first equation cannot become the second equation,
is that there are no \(\displaystyle \ x^2 \ terms\) in the second equation.**
\(\displaystyle ** \ \ (The \ \ x^2 \ term \ in \ the \ first \ equation \ has \ no \ other\)
\(\displaystyle x^2 \ terms \ to \ cancel \ it \ out.)\)
Last edited:
Hello, MrJoe2000!
There must be a typo . . .
We have: .. . . . . ..\(\displaystyle \dfrac{x^2}{2} - 4x + y^2 - 3y \:=\:-10\)
Multiply by 4: .\(\displaystyle 2x^2 - 16x + 4y^2 - 12y \:=\:-40\)
. . \(\displaystyle 2(x^2 - 8x \qquad) + (4y^2 - 12y \qquad) \:=\:-40\)
. . \(\displaystyle 2(x^2 - 8x + 16) + (4y^2 - 12y + 9) \:=\:-40 + 32 + 9\)
. . . . . . . . . . . .\(\displaystyle 2(x-4)^2 + (2y-3)^2 \:=\:1\)
There must be a typo . . .
\(\displaystyle \text{Can someone tell me how: }\:y^2 - 3y \:=\:4x - \frac{x^2}{2} - 10\)
. . . . . . . . . . . . . . . . \(\displaystyle \downarrow\)
\(\displaystyle \text{becomes: }\:2(x - 4)^2 + (2y-3)^2 \:=\:1\quad\text{ if it does at all?}\)
We have: .. . . . . ..\(\displaystyle \dfrac{x^2}{2} - 4x + y^2 - 3y \:=\:-10\)
Multiply by 4: .\(\displaystyle 2x^2 - 16x + 4y^2 - 12y \:=\:-40\)
. . \(\displaystyle 2(x^2 - 8x \qquad) + (4y^2 - 12y \qquad) \:=\:-40\)
. . \(\displaystyle 2(x^2 - 8x + 16) + (4y^2 - 12y + 9) \:=\:-40 + 32 + 9\)
. . . . . . . . . . . .\(\displaystyle 2(x-4)^2 + (2y-3)^2 \:=\:1\)