find the exact solution (s) of each system of equations. y=x^2+1 and y=2x?

Question

a. (1,2), (-1,2)
b. (-1,2)
c. (1,2)
d. (-1,2), (0,2)

Answer 1

since they're both solved for y, equate the values of y

2x = x²+1
0 = x²-2x+1
0 = (x-1)²
Double root, x = 1

y = 2x
(1,2)

Answer 2

The answer is c (which is (1,2)).
Set the two equations equal to each other because they are equal to the same variable (y).
x^2+1=2x
This can be rewritten x^2-2x+1=0
When this is factored, you will find that there is only one factor appearing twice: (x-1)
So the value of x is 1. (1-1=0 and you want the factor to equal 0).
Knowing that x is 1, substitute that value for x in the equation y=1^2 +1. That is calculated and the result is 2.
So x is 1 and y is 2.
Answer: (1,2) which is choice c.

Answer 3

(What handed off to #a million?) (2) No answer, simply by fact y can no longer in all probability be equivalent to the two 2x AND equivalent to 4 under 2x. (3) y = 2x and y - 2x = 0 are actually the comparable equation. and because there are infintely many recommendations to y = 2x, the respond is infinitely many recommendations. (4) there is in basic terms one answer, simply by fact if y = 2x, then you could subtitute 2x for y in the 1st equation and you get 2x = -3x + 2. yet that equation has in basic terms one answer for x (it somewhat is two/5), and for this reason y must be 2(2/5), or 4/5.

Answer 4

By substitution x^2-2x+1=(x-1)^2 = 0 so x=1 and y =2
yout answer is c)

Answer 5

set y1 = y2

x^2+1=2x ==> x^2-2x+1 = 0

quadratically

(x-1)(x-1) = 0

x1 = x2 =1

sub back into either equation

y = 1^2+1=2

therfore

(1,2)