Task 2—non-linear systems of equations create a system of equations that includes one linear equation and one quadratic equation. part 1. show all work to solving your system of equations algebraically. part 2. graph your system of equations, and show the solution graphically to verify your solution.

There are three parts. Complete answers for the three parts are below.

1) System of linear equations:

2) Algebraic solution:

3) Graph attaches showing intersection points (-1,1) and (2,4)

Step-by-step explanation:

1. Start: Create a system of equations that includes one linear equation and one quadratic equation.

i) Linear equation:

A linear equation is a first degree equation, i.e. it has the variables raised to the power 1.

The linear equations can be written in the form y = ax+b, where a and b are real numbers and y and x are the variables.

So, take this for our example: y = x + 2

ii) Quadratic equation

A quadratic equation is a second degree equation; the general form of a quadratic equation is y = ax² + bx + c, where a, b and c are real numbers, and x and y are the variables. In this, a cannot be zero.

Hence, take this for our example: y = x²

iii) System:

Part 1. Show all work to solving your system of equations algebraically.

1. Since, both equations in our system are equal to y, you start doing them both equal:

2. Apply equation properties to solve for x:

        x - 2 = 0 ⇒ x = 2

        x + 1 = 0 ⇒ x = - 1

Conclusion: the algebraic solutions are:

Part 2. Graph your system of equations, and show the solution graphically to verify your solution.

i) Both equations have to be graphed in one same coordinate system, and the solution will be the intersection points of the both graphs.

ii) To graph the linear equation you just need two points. The x and y - intercepts are usually excellent for this. This table show them:

       x      y = x + 2

      -2      - 2 + 2 = 0    ⇒point (-2, 0)

       0       y = 0 + 2 = 2 ⇒ point (0, 2)

iii) Graphing the quadratic equation normally requires more information. These are some guidelines you can use:

        y = A (x - h)² + k, where A is the leading coefficient, h is the      

        x-coordinate of the vertex and k is the y-coordinate of the vertex.

       Our equation, y = x² is equal to y = 1 (x - 0)² + 0, so you get:

        A = 1, h = 0, k = 0 ⇒ the vertex is the point (0,0).

         x       y = x²

        -5      (-5)² = 25

        -4      (-4)² = 16

        -3      (-3)² = 9

        -2      (-2)² = 4

        -1       (-1)² = 1

         0       0

         1         1

         2        4

         3        9

         4       16

         5       25

Such solution is attached. You can see that the two graphs intersect at (2,4) and (-1,1).

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.