(please show all work this helps me understand how you got the answers, thank you!)
Given set of equations are
\(y=x^2-3x+4\) ...(1)
and \(x+y=4\) ⇒ \(y=4-x\) ...(2)
From eq. (1) and (2)
∴ \(x^2-3x+4=4-x\)
\(x^2-2x=0\)
\(x(x-2)=0\)
\(x=0\) or \(2\)
For x=0 in eq (1) For x=2 in eq(1)
⇒ y = 4 ⇒ y = 2
∴ The solutions are (0,4) and (2,2).
~Looking forward