How do you solve #|8r+1|=3r# and find any extraneous solutions?

1 Answer
Mar 14, 2017

Answer:

There are no valid solutions to the given equation.

Explanation:

If #abs(8r+1) = 3r#
then
#{:("either ",8r+1 = 3r," or ",8r+1 = -3r), (,rarr 5r+1=0,,rarr 11r+1=0), (,rarr r=-1/5,,rarr r=-1/11) :}#

Note that neither of these solutions an be valid
since the left side must be positive and these solutions cause the right side to be negative.

However, we can check each possible solution explicitly:
[1]#color(white)("XXX")"if "r=-1/5 " then"#
#color(white)("XXX")LS#
#color(white)("XXXXX")=abs(8r+1)#
#color(white)("XXXXX")=abs(8 * (-1/5)+1)#
#color(white)("XXXXX")=abs(-3/5)#
#color(white)("XXX")RS#
#color(white)("XXXXX")=3r#
#color(white)("XXXXX")=3 * (-1/5)#
#color(white)("XXXXX")=-3/5#
#color(white)("XXX")LS != RS color(white)("XXX")#solution is extraneous.

[2]#color(white)("XXX")"if "r=-1/11" then"#
#color(white)("XXX")LS#
#color(white)("XXXXX")=abs(8r+1)#
#color(white)("XXXXX")=abs(8 * (-1/11)+1)#
#color(white)("XXXXX")=abs(3/11)#
#color(white)("XXXXX")=3/11#
#color(white)("XXX")RS#
#color(white)("XXXXX")=3r#
#color(white)("XXXXX")=3 * (-1/11)#
#color(white)("XXXXX")=-3/11#
#color(white)("XXX")LS!=RScolor(white)("XXX")#solution is extraneous

Here is the graph of #abs(8r+1)#
graph{abs(8x+1) [-1.107, 1.594, -0.085, 1.265]}

and here is the graph of #3r#
graph{3x [-0.6154, 1.2815, -0.3, 0.6484]}

Notice the lines for these two expressions have no points in common. (Sorry I can't remember how to get both of these on the same graph and I'm not using my normal computer with s/w to do these better).