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

Question

Abigail Allen · Accepted Answer

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.

Kennedy Adams · Answer

#X = 1/10 or x = -1/2# #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#

Avery Alexander · Answer

#x=-1/2" or "x=1/10# #"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)|)))#

Eli Assouline · Answer

undefined To solve the equation $2|$5x + 1$| - 3 = 0$, first, isolate the absolute value term by adding 3 to both sides:

$$2|$5x + 1$| = 3$$

Next, divide both sides by 2:

$$|$5x + 1$| = $\frac{3}{2}$$$

Now, since the absolute value can equal either the positive or negative value of its argument, set up two equations:

$$5x + 1 = \frac{3}{2}$$
$$5x + 1 = -\frac{3}{2}$$

Solve each equation separately:

For $5x + 1 = \frac{3}{2}$:

$$5x = \frac{3}{2} - 1$$
$$5x = \frac{1}{2}$$
$$x = \frac{1}{10}$$

For $5x + 1 = -\frac{3}{2}$:

$$5x = -\frac{3}{2} - 1$$
$$5x = -\frac{5}{2}$$
$$x = -\frac{1}{2}$$

So, the solutions to the equation are $x = \frac{1}{10}$ and $x = -\frac{1}{2}$.