Solve the following system of equations and show all work. y = x2 + 3 y = x + 5
giving 20 points, .

We will see that the solutions of the system of equations are (-2, 3) and (1, 6).

Solving the system of equations:

So we have the system:

y = x^2 + 3

y = x + 5

We can see that the variable y is already isolated in both equations, then we can write:

x^2 + 3 = y = x + 5

x^2 + 3 = x + 5

Now we can solve the above equation for x, we will get:

x^2 + 3 - x - 5 = 0

x^2 - x - 2 = 0

This is a quadratic equation, the solutions are given by:

Then the two solutions are:

To get the y-values we just replace these in one of the equations, we will get:

for x = -2

y = x + 5 = -2 + 5 = 3

Then the point (-2, 3) is a solution of the system.

for x = 1

y = x + 5 = 1 + 5 = 6

So the point (1, 6) is a solution of the system.

If you want to learn more about systems of equations, you can read:

link

(-1, 4) and (2, 7)

Step-by-step explanation:

If both of those equations equal y, the we can set them equal to each other and solve for x.  There will be 2 solutions because this is a second degree polynomial.

Get everything on one side of the equals sign, set it equal to 0, then factor to solve for x.

If you plug these values into the quadratic formula you will get x values of -1 and 2.  We first plug -1 into either one of the original equations to solve for y:

y = (-1) + 5 so

y = 4 and the resulting coordinate is (-1, 4)

Plug in 2 into the same equation to find y:

y = 2 + 5 so

y = 7

a. the residual plot for model 1 has a random pattern and is a good fit for the data.

24 units

step-by-step explanation: