Question

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

Answer 1

`x=1/3" or "x=-5/3`

Explanation:

`"distribute the factor"`

`rArr|(2(1/3+1/2x)|=|2/3+x|`

`"the value inside the absolute value function can be"`
`color(blue)"positive or negative"`

`color(blue)"first solution"`

`2/3+x=1rArrx=1-2/3=1/3`

`color(blue)"second solution"`

`-(2/3+x)=1`

`rArr-2/3-x=1`

`rArr-x=1+2/3=5/3rArrx=-5/3`

`color(blue)"As a check"`

`x=1/3"`

`rArr|2/3+1/3|=|1|=1`

`x=-5/3`

`rArr|2/3-5/3|=|-1|=1`

Answer 2

Use the piecewise definition of the absolute value function to separate the equation into two equations and then solve each equation.

Explanation:

The piecewise definition of the absolute value function is:

`|f(x)| = {(f(x); f(x) >= 0),(-f(x); f(x) < 0):}`

In this case `f(x) = 2(1/3+1/2x)`

Substitute into the definition:

`|2(1/3+1/2x)| = {(2(1/3+1/2x); 2(1/3+1/2x) >= 0),(-2(1/3+1/2x); 2(1/3+1/2x) < 0):}`

Simplify the domain restrictions:

`|2(1/3+1/2x)| = {(2(1/3+1/2x); x >= -2/3),(-2(1/3+1/2x); x < -2/3):}`

Separate the given equation into two equations with its respective domain restriction:

`2(1/3+1/2x) = 1; x >=-2/3` and `-2(1/3+1/2x) = 1; x < -2/3`

Multiply the second equation by -1:

`2(1/3+1/2x) = 1; x >=-2/3` and `2(1/3+1/2x) = -1; x < -2/3`

Distribute the two in both equations:

`2/3+x = 1; x >=-2/3` and `2/3+x = -1; x < -2/3`

Subtract `2/3` from both sides of both equations:

`x = 1/3; x >=-2/3` and `x = -5/3; x < -2/3`

The domain restrictions can be dropped, because neither equation violates them:

`x = 1/3` and `x = -5/3`