Solving a Mixed-Degree System of Equations
Consider the following system of equations:
10 + y = 5x + x2
5x + y = 1
The first equation is an equation of a
The second equation is an equation line
DONE
A ✩
parabola
circle

Respuesta :

Answer:

  • The first equation is of a parabola
  • (x, y) = (-11, 56) or (1, -4)

Step-by-step explanation:

You want a description and solution of the system of equations ...

  • 10 +y = 5x +x²
  • 5x +y = 1

Description

The first equation is 2nd degree in x and first degree in y. It is the equation of a parabola.

Solution

We can add 5x to the first equation to get ...

  5x +y +10 = 10x +x²

Using the second equation to replace (5x +y), we have ...

  1 +10 = 10x +x²

  x² +10x -11 = 0 . . . . . . . subtract 11 to put in standard form

  (x +11)(x -1) = 0 . . . . . . factor

The solutions are the values of x that make these factors zero:

  x = -11, or x = 1

The corresponding values of y are ...

  y = 1 -5x

  y = 1 -5(-11) = 56

  y = 1 -5(1) = -4

The solutions are (-11, 56) and (1, -4).

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Additional comment

We find it convenient to use a graphing calculator to solve systems of equations like this. Even when the solutions are fractions or irrational numbers, the graph can be useful.

For a system like this, graphing the function f(x) = x² +5x -11 and identifying its x-intercepts will give you the x-values of the solutions. Then the y-values can be found using either of the original equations.

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