solve the system of equations.
@kirbykirby
he is great he can help you, not only so you get the answer but you understand it
you know what y is so substitute it into the other equation and simplify to work out the x value, then substitute x back into the first equation
for example
\[y= 2x ^{2} -3\] \[y=3x-1\] We know Y so \[2x ^{2}-3=3x-1\]
i think its a but im not sure. i just need someone to tell me if im wrong or not.
then you just solve the quadratic, get the x values, and then sub your x values into one of the original equations to get your y values
so its b?
no wait..
I sort of ignored the "no solution" answer
what did you find as your x-values?
i was thinking it was no solution because i worked the problem out. i just wanted someone to double check.
Well there is a solution =]
my other thought would be B.
b is correct isn't it?
Well not quite. Just substitute \(x=-1/2\) and you don't get y=5
so which is it? @kirbykirby
oh I didn't see "no solution" lmao
the answer will be (-1/2, -5/2) (2,5)
I did the same mistake not seeing "no solution" at first xD @amorfide But yeah you're right =]
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