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:
Its roots are given by:
We have two roots:
Thus, we look for the values of y, by substituting any of the equations:
(4,6) or (-1,-4)
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
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
It's D, (5, 4).
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
Option D. (5,4)
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:
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:
Solution: x=5 and y=4: Point=(x,y)→Point=(5,4)
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)