Question

How do you solve the system of equations `C-4r=-3` and `2C-8r=-6`?

Answer 1

The lines are the same, so there is an `"infinite number"` of solutions.

Explanation:

`color(white)(.)`C - 4r = - 3
2C - 8r = - 6

You can give the equations the same coefficient for C by multiplying the first equation by 2.

After you do that, this is the system of equations:
2C - 8r = - 6
2C - 8r = - 6

In other words, both equations are for the same line.

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . .

When you are graphing systems of equations, there are a few kinds of systems:

• The lines may intersect at exactly one point.
  That point is `"the solution"` to the system.

http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L1_T1_text_final.html

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . .

• If the lines are parallel, they never intersect.
  So there are `"no solutions"` to that system.

http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L1_T1_text_final.html

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . .

• The lines are the same line.
  They intersect at every point, so there are an `"infinite number"` of solutions.
  
  http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L1_T1_text_final.html

`color(white)(mmmmmmmm)` . . . . . . . . . . . . . .

Here's a web site where you can find out more about systems of equations:
http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U14_L1_T1_text_final.html