we have
y=x^2-9x+18 -----> equation A
y=x-3 ------> equation B
Solve the system of equations
substitute equation B in equation A
x^2-9x+18=x-3
x^2-9x+18-x+3=0
x^2-10x+21=0
Solve the quadratic equation using the formula
[tex]x=\frac{-b\pm\sqrt[\square]{b^2-4ac}}{2a}[/tex]
we have
a=1
b=-10
c=21
substitute the given values
[tex]\begin{gathered} x=\frac{-(-10)\pm\sqrt[\square]{(-10)^2-4(1)(21)}}{2(1)} \\ \\ x=\frac{10\pm\sqrt[\square]{100-84}}{2} \\ \\ x=\frac{10\pm\sqrt[\square]{16}}{2} \\ \\ x=\frac{10\pm4}{2} \\ \\ x=7 \\ x=3 \end{gathered}[/tex]
Find the value of y for x=7
y=x-3
y=7-3=4
the first solution is (7,4)
Find the value of y for x=3
y=3-3=0
the second solution is (3,0)
therefore
the answer is the first option
