Question

Solve the following system algebraically. y= x2 - 9x + 18 y = x - 3

Answer 1

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

`x=\frac{-b\pm\sqrt[\square]{b^2-4ac}}{2a}`

we have

a=1

b=-10

c=21

substitute the given values

`\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=(10\pm4)/(2) \\ \\ x=7 \\ x=3 \end{gathered}`

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