Your post is not displaying correctly (at least for me).
What kind of code is that?
Is that supposed to be LaTeX?
thth776 is taking a vacation.
Still, if anyone knows how to start this I'd appreciate the hint. I'm kinda curious how to go about this one myself.
Edit: Silly me. Just graph the thing. No worries then.
There could be a different approach. Let us see if get a different idea. In the mean time have this
Hint: recollect the definition of modulus function, |x| = x for all x >= 0 and |x| = -x for all x <0.
For second case first go by the assumption that 3x-2y+10 >0 and 2x-3y>0, then proceed by solving with elimination by substitution.
In other words, because 3x-2y+10 and 2x-3y are expressed as absolute-value functions, we have to assume both for when they are positive, and for when they are negative.
Because the two absolute value functions are being added together, they are commutative, and four possibilities will arise. Two of which will have real-valued solutions.
One of them is easy enough to solve.
The other requires solving a quartic function.
Sure. When you think about absolute value functions, by definition, we know that the outcome is always positive no matter whether we have a negative or positive value on the inside.
|3| = 3 and |-3| = 3.
It didn't matter whether it was a 3 or a -3, the outcome is a positive 3.
So, if we turned things around and wanted to solve for an unknown variable in an absolute value function:
|x| = ?
How would we know whether the x was a positive or negative value? Regardless of whether it is positive or negative, the outcome is going to be the same—it will be a positive outcome. So because they share the same outcome, we consider both the positive and negative value as solutions of the absolute-value function that are both valid. In this latter example, it would mean that we have to consider BOTH x and -x as solutions.
How does this connect with our problem in this post?
When we work with analytic expressions instead of numbers in absolute-value functions, we don't treat them differently.
So, with , we have to consider two things:
This would mean that for , you would discover that there would be four equations. Think about what was shown earlier, and see how it relates with this problem. With these four equations, we have a system of equations with .
You would have to solve from here—seeing where the four equations intersect with .