Question

Solve the following quadratic-linear system of equations.

Answer 1

D.
`(5,4)`

Explanation:

The given quadratic linear system of equations is:

`y=x^2-8x+19...eqn(1)`.

`y=2x-6...eqn(2)`

We equate the two equations to obtain;

`x^2-8x+19=2x-6`.

Rewrite in the general quadratic form to get;

`x^2-8x-2x+19+6=0`

`\Rightarrow x^2-10x+25=0`.

Observe that the expression on the left hand side is a perfect quadratic trinomial.

`\Rightarrow (x-5)^2=0`.

`\Rightarrow x-5=0`.

`\Rightarrow x=5`

Substitute
`x=5` into either equation (1) or (2) to get;

`y=2(5)-6`

`y=10-6`

`y=4`

The solution is
`(5,4)`.

Therefore the correct answer is D

Answer 2

Option D. (5,4)

Explanation:

`y=x^(2)-8x+19\\ y=2x-6`

Using the method of equaling the two equations:

`y=y\\ x^(2)-8x+19=2x-6`

This is a quadratic equation, then we must equal to zero. Equaling to zero subtracting 2x and adding 6 to both sides of the equation:

`x^(2)-8x+19-2x+6=2x-6-2x+6\\ x^(2)-10x+25=0`

Factoring:

`(x-5)(x-5)=0\\ (x-5)^(2)=0`

Solving for x: Square root both sides of the equation:

`\sqrt{(x-5)^(2) }=√(0)\\ x-5=0`

Adding 5 to both sides of the equation:

`x-5+5=0+5\\ x=5`

Replacing x=5 in any of the two given equations:

y=2x-6

y=2(5)-6

y=10-6

y=4

Solution: x=5 and y=4: Point=(x,y)→Point=(5,4)