Question

55:06 Which represents the solution(s) of the graphed system of equations, y = –x2 + x + 2 and y = –x + 3? (1, 2) (1, 2) and (0, 3) (–1, 0) and (2, 0) (2, 1)

Answer 1

Answer:

(1,2)

Step-by-step explanation:

We have the system: $\left \{ {{y=-x^2+x+2} \atop {y=-x+3}} \right.$ to see which points represents the solution(s) of the system we have to replace them in both equations and see if the points verify them.

Point (1,2):

x=1, y=2

First equation:

$y=-x^2+x+2\\2=-(1)^2+1+2\\2=-1+1+2\\2=2$

Is verified.

Second equation:

$y=-x+3\\2=-1+3\\2=2$

Is verified.

Then this is the correct answer.

Points (1,2) and (0,3):

We saw that the point (1,2) verified both equations.

Now we are going to replace (0,3) in both equations:

$y=-x^2+x+2\\3=-(0)^2+0+2\\3\neq2$

It isn't verified for the first equation.

$y=-x+3\\3=-0+3\\3=3$

It is verified for the second.

Then this is not the solution of the system because (0,3) belongs only to the second equation.

Points (-1,0) and (2,0):

We have to replace (-1,0) in both equations:

First equation:

$y=-x^2+x+2\\0=-(-1)^2-1+2\\0=-2+2\\0=0$

Is verified.

Second equation:

$y=-x+3\\0=-1+3\\0\neq2$

It is not verified for the second equation. The point (-1,0) belongs to the graph of the first equation only.

We are going to replace (2,0) in the first equation:

$y=-x^2+x+2\\0=-(2)^2+2+2\\0=-4+4\\0=0$

Is verified.

Now in the second equation:

$y=-x+3\\0=-2+3\\0\neq1$

It isn't verified for the second equation. The point (2,0) belongs to the first equation.

Then this doesn't represent the solution.

Point (2,1):

First equation:

$y=-x^2+x+2\\1=-(2)^2+2+2\\1=-4+4\\1\neq2$

It isn't verified for the first equation.

Second equation:

$y=-x+3\\1=-2+3\\1=1$

Is verified.

Then this answer is incorrect, because the point is verified only for the second equation.

The graph of the system is attached, where $y=-x^2+x+2$ is the blue one and $y=-x+3$ the pink one.

Answer 2

Hi.
This is not hard.Just plug in the values.
(1, 2) is a solution. It gives the both equations a value of 2, when given a 1
(0,3) is not a solution. The first equation returns a 2, not a 3. The 0 pairs for C are not correct either, since the second is not a quadratic.
(2,1) is not a solution. Since an even number is being put in the first equation, there is no chance of the answer being odd, since you are adding all even numbers
Hope this helps.