How do you solve the equation #2abs(5x+1)-3=0#?

Given: #2|5x+1|-3=0#

Add 3 to both sides:

#2|5x+1|=3#

Divide both sides by 2

#|5x+1| = 3/2#

Separate into two equations without the absolute value function, one with the right side positive and the other with the right side negative:

#5x+1 = 3/2# and #5x+1 = -3/2#

Subtract one from both sides of both equations:

#5x = 1/2# and #5x = -5/2#

Divide both sides of both equations by 5:

#x = 1/10# and #x = -1/2#

Check:

#2|5(1/10)+1|-3=0# and #2|5(-1/2)+1|-3=0#

#2|3/2|-3=0# and #2|-3/2|-3=0#

#3-3=0# and #3-3=0#

Both values check.

#2|5x + 1| - 3 = 0#
#2|5x + 1| = 0 + 3 #
#2|5x + 1| = 3#
#|5x +1| = 3/2#
We know either
#5x + 1 = 3/2 or 5x +1 = -3/2#
Let's solve the first one
#5x + 1 = 3/2#
#5x = 3/2 - 1#
#5x = 1/2#
Divide both sides by 5
#5x/5 = 1/2/5#
#x = 1/10#
Solve the second one
#5x + 1 = -3/2#
#5x = -3/2 -1#
#5x = -5/2#
Divide both sides by 5
#5x/5 = -5/2/5#
#x = -1/2#

Therefore,
#x = 1/10 or x = -1/2#

#"isolate the absolute value"#

#"add 3 to both sides"#

#2|5x+1|cancel(-3)cancel(+3)=0+3#

#rArr2|5x+1|=3#

#"divide both sides by 2"#

#cancel(2)/cancel(2)|5x+1|=3/2#

#rArr|5x+1|=3/2#

#"the expression inside the absolute value can be"#
#"positive or negative"#

#color(blue)"Solution 1"#

#5x+1=3/2#

#"subtract 1 from both sides"#

#rArr5x=1/2#

#"dividing both sides by 5 gives"#

#rArrcolor(red)(bar(ul(|color(white)(2/2)color(black)(x=1/10)color(white)(2/2)|)))#

#color(blue)"Solution 2"#

#-(5x+1)=3/2#

#rArr-5x-1=3/2#

#"add 1 to both sides"#

#rArr-5x=5/2#

#"divide both sides by - 5"#

#rArrcolor(red)(bar(ul(|color(white)(2/2)color(black)(x=-1/2)color(white)(2/2)|)))#