Question

Solve the system of equations. Y=2x^2-3 Y=3x-1
A. No solution
B. (-1/2, 5), (2, -5/2)
C. (-1/2, -5/2), (2, 5)
D, (1/2, 5/2) , (2, 5)

Answer 1

\(\large\color{black}{ \displaystyle y=2x^2-3}\)
\(\large\color{black}{ \displaystyle y=3x-1}\)


in the first equation the 'y' is equivalent to '2x^2-3' and in the second equation (this same y' is equivalent to) '3x-1'

so wouldn't that mean that 3x-1=2x^2-3 ?

Answer 2

basically, given \(\large\color{black}{ \displaystyle a=b}\) and \(\large\color{black}{ \displaystyle a=c}\)
therefore \(\large\color{blue}{ \displaystyle b=c}\).

Answer 3

here, we go by the same exact principle, but the terms are not a,b,c.

Answer 4

So, based on what I said, what do you think you should do?

Answer 5

I'm so confused.. :(

Answer 6

lol he basically blew up your brain! :'D I believe it's A no solution because n a linear equation and quadratic equation cannot come together :)

Answer 7

That's what I was thinking but I just got really confused. Thank you.

Answer 8

your welcome :)

Answer 9

that is not true at all !!

Answer 10

I bet 99.999999 % there is a solution at least 1.

Answer 11

Anyway,
do you understand the a=b a=c and therefore b=c
principle that I posted ?

Answer 12

Yeah. I read it over a couple of times.

Answer 13

ok, so what you should then say is:
\(\large\color{black}{ \displaystyle 2x^2-3=3x-1 }\)

Answer 14

Then, solve this quadratic