Question

quadratic system

2x^2 = 4y

2x-2y = -5

please show the work with the answer, i want to know how to do quadratic systems. elimination method if possible, please

Answer 1

\[2x^2 = 4y..............(1)\]
\[2x-2y = -5 ...............(2)\]
--------------------------------
\[2x^2 = 4y..................(1)\]
\[2x= -5+2y ..........(2)\]
Square both sides,
\[2x^2=4y.............(1)\]
\[4x^2=(-5+2y)^2.............(2)\]
From (2),
\[2x^2=\frac{(2y-5)^2}{2}........(2)\]
------------------------------------
\[2x^2 = 4y..................(1)\]
\[2x^2=\frac{(2y-5)^2}{2}........(2)\]
Elimination
\[0=4y-\frac{(2y-5)^2}{2}\]
\[y\text{ = }\frac{1}{2} \left(7\pm2 \sqrt{6}\right)\]
Find x using the y valaues
As you can see elimination method here is more complicated than substitution method.

Answer 2

@.Sam. we do not need sites as we have a very good helper like u :)
Very nice solution :)

Answer 3

it boggles me though @.Sam. ....wouldnt it have been easier to cancel y rather than x? i mean put the equations in the form y =