Question

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

Answer 1

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.

Answer 2

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

Explanation:

`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`

Answer 3

`x=-1/2" or "x=1/10`

Explanation:

`"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)|)))`