Question

A system of a linear and quadratic equation is shown below.

Answer 1

Answer:

The solutions are the points (1,4) and (4,1)

Step-by-step explanation:

we have

$6x+y=x^{2}+9$ ----> equation A

$x+y=5$ ----> $y=-x+5$ ---> equation B

substitute equation B in equation A

$6x+(-x+5)=x^{2}+9$

solve for x

$6x-x+5=x^{2}+9$

$5x+5=x^{2}+9$

$x^{2}+9-5x-5=0$

$x^{2}-5x+4=0$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}-5x+4=0$  

so

$a=1\\b=-5\\c=4$

substitute in the formula

$x=\frac{-(-5)(+/-)\sqrt{-5^{2}-4(1)(4)}} {2(1)}$

$x=\frac{5(+/-)\sqrt{9}} {2}$

$x=\frac{5(+/-)3} {2}$

$x=\frac{5(+)3} {2}=4$

$x=\frac{5(-)3} {2}=1$

The solutions are x=1,x=4

Find the values of y

For x=1 ----->  $y=-(1)+5=4$  ---->  (1,4)

For x=4 ----->  $y=-(4)+5=1$  ---->  (4,1)

therefore

The solutions are the points (1,4) and (4,1)