Question

How do you solve `\frac { 3} { 5} | 10- 5x | + 7> 25`?

Answer 1

when `x<-4` or `x>8`

Explanation:

the first step is to simplify the expression as much as we can.
`3/5 |10-5x| +7 >25`

`3/5 |10-5x| >25-7`

`|10-5x| > (18*5)/3`

`|10-5x| > 30`

Now `|10-5x|` contains two different functions the f(x) when f(x) is positive or -f(x) when f(x) is negative.

the turning point can easily be found be equating to zero,

`10-5x=0`

`x=2`

so when x is less than 2 the function is `10-5x` and when x is greater than two the function is `-10+5x`

so to find the values that make all these conditions true we must equate.

`10-5x>30`

`-5x>20`

`x<-4`
so, check that this x value meets the requirement of being less than 2. (which it does)

and

`-10+5x>30`

`5x>40`

`x>8`
so, check that this x value meets the requirement of being greater than 2. (which it does)

so we have our two solutions.
This is true either when x is less then -4 or greater than 8.

it is easier if you visualise it as a graph,
graph{|10-5x| [-30.65, 34.3, 8.55, 41.05]}

Answer 2

`x<-4` , `x>8`

Explanation:

Another way to solve this is using an algebraic technique.

`3/5|10-5x|+7>25`

`|10-5x|>30`

`|10-5x|>30` can be rewritten as `sqrt((10-5x)^2)>30`
on expansion,

`10-5x> +-sqrt(30^2)`

`10-5x>30` , `10-5x> -30`

`-5x>20` , `-5x> -40`

`x<-4` , `x>8`

(both sides are negative so no change of sign is needed for `x>8`)