Question

Solve the following quadratic-linear system of equations.

Answer 1

Answer:

Option D. (5,4)

Step-by-step 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:

$\sqrt{(x-5)^{2} }=\sqrt{0}\\ x-5=0$

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)

Answer 2

Answer:

D. $(5,4)$

Step-by-step 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