Answer:
[tex]|3r-15| = -3r+15[/tex] if [tex]r<5[/tex]
Step-by-step explanation:
Given that:
r < 5
To simplify:
|3r−15|
Solution:
First of all, let us learn about Modulus function:
[tex]f(x) =|x| =\left \{ {x, \ if\ {x>0} \atop -x\ if\ {x<0}} \right.[/tex]
In other words, we can say:
Modulus function has a role to make its contents positive.
If the contents are positive, the result will be equal to its contents only.
If the contents are negative, it will add a negative sign to the contents to make it positive.
Now, let us consider the given condition:
[tex]r < 5[/tex]
Multiply both sides with 3. (As 3 is a positive number, the equality sign will not change.)
[tex]3r < 15[/tex]
Subtracting 15 from both sides:
[tex]3r-15<15-15\\\Rightarrow 3r-15<0[/tex]
Now, we know that [tex]3r-15<0[/tex], let us use the definition of Modulus function.
Add a negative sign to the contents because the contents are already negative.
[tex]\\\Rightarrow |3r-15| = -(3r-15)\\\Rightarrow -3r+15[/tex]
So, the answer is:
[tex]|3r-15| = -3r+15[/tex] if [tex]r<5[/tex]
