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

1 Answer
Jan 26, 2018

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.

enter image source here
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.

enter image source here

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

#color(white)(mmmmmmmm)# . . . . . . . . . . . . . .

#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