Question

How do you solve `|x – 4| > |3x – 1|`?

Answer 1

`x inRR`

Explanation:

When there is absolute value, we have to take two options the first where the first is positive the next is negative.
`|a+b|`
`=a+b`
or `=-(a+b)=-a-b`
`|x–4|>|3x–1|`

1. `=x-4>|3x-1|`
again doing the same with the left hand side:
`x-4>|3x-1|`

1a. `=x-4>3x-1`

Subtracting x from both sides of the equation to get the unknown's terms to one side of the equation:
`cancelxcancel(-x)-4> 3x-1-x`
`-4> 2x-1`

Adding 1 to both sides of the equation to isolate the unknown's term on the left hand side:
`-4+1> 2xcancel(-1)cancel(+1)`
`-3> 2x`

Dividing both sides of the equation by 2 to isolate x:
`-3/2> (cancel2 x)/(cancel2)`

1a. `color(red)(x<-3/2`

and

1b. `x-4 > -(3x-1)`
`=x-4> -3x+1`
Doing same as before to isolate x:
`cancelx-4cancel(-x)> -3x+1-x`
`-4> -4x`
`(cancel(-4))/cancel(-4)> (cancel(-4)x)/cancel(-4)`
`1>x`
1b. `color(red)(x<1`

and

1. `=-(x-4)>|3x-1|`
   `=-x+4>|3x-1|`

2a.`=-x+4>3x-1`

simplifying: (adding 1 and subtracting x from both sides of the equation and then dividing by 4):
`cancel(-x)cancel(+x)+4+1>3xcancel(-1)+xcancel(+1)`
`4+1>3x+x`
`5>4x`

`5/4>(cancel4 x)/(cancel4)`
`5/4>x`
2a.`color(red)(x<5/4`

and

2b.`=-x+4> -1(3x-1)`
`=-x+4> -3x+1`

Simplifying by adding 3x to, subtracting 4 from both sides and then dividing by 2:
`-xcancel(+4)+3xcancel(-4)> cancel(-3x)+1cancel(+3x)-4`
`-x+3x>1-4`
`2x>-3`
`(cancel2x)/cancel 2> -3/2`
2b. `color(red)(x> -3/2)`

Simplifying 1a , 2a ,and 2b :
`x< -3/2`
`x<1`
`x<5/4rarrx<1.25`
`rarr x<-3/2<1<5/4`
`rarr color(blue)(x<5/4`

and

From 2a:
`color(blue)(x> -3/2`

since x is smaller than `5/4` and bigger than `-3/2` then x defined for any number.