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

Add 3 to both sides:

Divide both sides by 2

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

Subtract one from both sides of both equations:

Divide both sides of both equations by 5:

Check:

Both values check.

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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}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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