Solve the system of equations. Show your work.

\(\boxed{1}\hspace{3mm}y=x^2-3x+4\\ \boxed{2}\hspace{3mm}x+y=4\)

 

In the system of equations, the first equation is already solver for, so I can substitute its value into the second question. 

 

\(y=\textcolor{red}{x^2-3x+4};\\ x+\textcolor{red}{y}=4\)	Use substitution to get rid of one variable.
\(x+\textcolor{red}{x^2-3x+4}=4\)	Combine like terms and subtract 4 from both sides.
\(x^2-2x=0\)	Factor the GCF of the left-hand side of the equation: x.
\(x(x-2)=0\)	Set both factors equal to 0 and solve.
\(x_1=0\) \(x-2=0\)   \(x_2=2\)

 

Let's substitute both possibilities for y: 

 

\(\boxed{2}\hspace{3mm}x+y=4\)	Let's substitute in the first possibility for x: x=0.
\(0+y=4\)
\(y_1=4\)
\(\boxed{2}\hspace{3mm}x+y=4\)	Substitute for the second option: x=2.
\(2+y=4\)
\(y_2=2\)

 

Here are the answers in coordinate form: \((0,4)\) and \((2,2)\)