Solve the following quadratic-linear system of equations.

Option D. (5,4)

Step-by-step explanation:

Using the method of equaling the two equations:

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:

Factoring:

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

Adding 5 to both sides of the equation:

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)

D.

Step-by-step explanation:

The given quadratic linear system of equations is:

.

We equate the two equations to obtain;

.

Rewrite in the general quadratic form to get;

.

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

.

.

Substitute into either equation (1) or (2) to get;

The solution is .

Therefore the correct answer is D

(4,6) or (-1,-4)

Step-by-step explanation:

since both equations about x are equal to y, you can make x^2-x-6=2x-2

solve this quadratic equation, you get x=4 or -1

substitute x into y=2x-2

you get y=6 when x=4, or y= -4 when x=-1

For this case we have a system of two equations with two unknowns.

To solve, we equate both equations:

We have a quadratic equation of the form:

Where:

Its roots are given by:

We have two roots:

Thus, we look for the values of y, by substituting any of the equations:

Option B

It's D, (5, 4).

Step-by-step explanation:

Substitute the second equation  for y in the first equation:-

2x - 6 =  x^2 - 8x + 19

x^2 - 10x + 25 = 0

(x - 5)^2 = 0

x = 5 (multiplicity 2).

So y = 2(5) - 6 = 4

You can do substitution 2x-2=x^2 -x -6; isolate all terms on one side 0= x^2 -x-2x -6+2; combine like terms x^2 -3x -4=0; factor the quadratic (x-4)(x+1)=0; each term is zero x-4=0 so x=4 and x+1=0 so x=-1. Now, y=2•4-2=6 and y=2•(-1) -2= -4 ; solutions for the system are ( 4,6) and ( -1, -4)