Simultaneous quadratic & linear: 3x + 2y = 7, Y= x^2 - 2x + 3

[MathsFormula]

Having trouble solving the simultaneous equation :

3x + 2y = 7

Y= x² - 2x + 3


3x + 2(x² - 2x + 3) y = 7

3x + 4x²-4x + 6 = 7

4x² - x - 1 = 0


I'm trying to factorise but can't do it. Perhaps I need to use the quadratic solver equation?

Is it possible to factorise?

Book answer is x= 1 y = 2 or x = -1/2 y = 17/4

Please help

 

[Prove It]

Having trouble solving the simultaneous equation :

3x + 2y = 7

Y= x² - 2x + 3


3x + 2(x² - 2x + 3) y = 7 You have an extra y here for some strange reason...

3x + 4x²-4x + 6 = 7 Last time I checked, 2(x^2) was 2x^2, not 4x^2...

4x² - x - 1 = 0

Fixing up this mistake you should get

\(\displaystyle \displaystyle \begin{align*} 3\,x + 2\,x^2 - 4\,x + 6 &= 7 \\ 2\,x^2 - x + 6 &= 7 \\ 2\,x^2 - x - 1 &= 0 \\ 2\,x^2 - 2\,x + x - 1 &= 0 \\ 2\,x \, \left( x - 1 \right) + 1\,\left( x - 1 \right) &= 0 \\ \left( x - 1 \right) \left( 2\,x + 1 \right) &= 0 \end{align*}\)

You should be able to finish now...

 

[MathsFormula]

Thanks. Seen the mistake I made and will continue the sum. Think I'll be okay now. Will get back to you if I encounter difficulties