Question

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

6x+y=x^2+9
x+y=5
When solving the system algebraically, what is the solution?

Answer 1

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

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(+/-)√(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)