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

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Answer 1 · 2018-01-26T06:11:48

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

$.$ $color(white)(.)$ 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.

$m m m m m m m m$ $color(white)(mmmmmmmm)$ . . . . . . . . . . . . . .

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 $the solution$ $"the solution"$ to the system.

$m m m m m m m m$ $color(white)(mmmmmmmm)$ . . . . . . . . . . . . . .

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

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$m m m m m m m m$ $color(white)(mmmmmmmm)$ . . . . . . . . . . . . . .

The lines are the same line.
They intersect at every point, so there are an $infinite number$ $"infinite number"$ of solutions.

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

$m m m m m m m m$ $color(white)(mmmmmmmm)$ . . . . . . . . . . . . . .

How do you solve the system of equations C−4r=−3C-4r=-3 and 2C−8r=−62C-8r=-6?