What is the procedure to solve equations that have multiple expression in absolute values?
Are there popular methods to solve them?
I recently learned, $$|x|≠ ± x $$
So I'm confused as to how does one solve equation that has multiple expression in absolute values like
$$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$$
– Thanks :)
On
First, note the definition: $$|f(x)| = f(x) \ \ \ \ \text{if }\ \ \ \ f(x)\ge0\\|f(x)| = -f(x) \ \text{otherwise}$$ This, in very simple terms, means that any solution to $|f(x)| =$ something is either a solution to $f(x) =$ something, or a solution to $-f(x) =$ something.
so, for instance, in your case, you have: $$|f(x)|+|g(x)|-|h(x)|=5\\f(x)=3x+4\\g(x)=2x−1\\h(x)=\frac{x+5}{4}$$
A solution to this equation must also be a solution to one of these: $$f(x)+g(x)+h(x)=5\\f(x)+g(x)−h(x)=5\\f(x)-g(x)+h(x)=5\\f(x)-g(x)−h(x)=5\\-f(x)+g(x)+h(x)=5\\-f(x)+g(x)-h(x)=5\\-f(x)-g(x)+h(x)=5\\-f(x)-g(x)−h(x)=5$$
It is simple enough to find all of the solutions to each of these, and then check to see if each of these is a solution to the original equation (they may not be, but every solution to the original will be a solution to at least one of these, so you wont miss any). This requires a fair amount of work, and you can be cleverer than this, but I find this to be the most intuitive approach.
To reiterate, any time I have |f(x)| in an equation, I can find the solutions to the equation with |f(x)| replaced with f(x) and the equation with |f(x)| replaced with -f(x), and simply check those values to see if they are solutions in the equation with |f(x)|.
If you wish to delve a little bit deeper, you might notice that we do not have to check the solution against the original equation, but merely check that the solution lies in the set where the equations themselves are equal, this means solving the inequalities $$f(x) \ge 0 \\g(x)\ge 0 \\ h(x) \ge 0 \\f(x) \le 0 \\g(x)\le 0 \\ h(x) \le 0$$ we can then associate each of the set of 8 functions we created earlier with the set that satisfies the corresponding collection of inequalities. This would help us, for instance, in the case where, instead of solving $|f(x)|+|g(x)|-|h(x)|=5$ we were solving $|f(x)|+|g(x)|-|h(x)|>5$, where we have infinitely many solutions, and the only way to express them is as the intersection of the sets satisfying each condition separately.
On
To say "$|x| \ne \pm x$" is a little weird.
The problem isn't whether it is true or not but that what it actually means is not clear.
$|x| = x$ if $ x \ge 0$. And $|x| = -x$ if $x \le 0$. So it is true that $|x| =x$ or that $|x| = -x$.
But what does $ \pm x$ mean? First off, there is no single number that is $\pm x$. It's not like Schrodenger's cat, that there is some number that is both $x$ and also $-x$ both at the same time. I'd say $\pm x$ is simply shorthand for "we want to use one of the values of $x$ or of $-x$ but either we aren't sure which one or in general our result will apply to either so we won't distinguish which".
In that sense, I, personally, interpret a statemen $k = \pm x$ to mean "either $k = x$ or $k = -x$". Or alternatively I might take it to mean "$\pm x$ is the group of $\{x,-x\}$ and $k$ is one of those members: i.e. '$k=\pm x$' means '$k \in \{x, -x\}$".
So to me. I'd so $|x| = \pm x$ but not the Schrondenger cat superimposed single value idea but that eithe $|x| = x$ or $x= -x$.
....
So to do equations. I do cases: Case 1: $|x| = x$ and $x \ge 0$ and Case 2: $|x|= -x$ and $x < 0$.
For example:
$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$
$|3x+4| = \pm(3x + 4); |2x-1|=\pm (2x-1); |x+5| = \pm 5$. There are three chooses each with two options. That means there are $2^3 = 8$ cases.
That may seem like a lot but that is okay. Some of them will be self contradictory and some will be redundent.
Case 1:
$|3x + 4| = 3x+4$ because $3x + 4 \ge 0$ and $|2x-1| = 2x -1$ because $2x -1 \ge 0$ and $|x+5| = x+5$ because $x + 5 \ge 0$.
Now before I even start I want to clear up my conditions for this case.
$3x + 4 \ge 0$ so $x \ge \frac 43$. And $2x -1 \ge 0$ so $x \ge \frac 12$. And $x+5 \ge 0$ so $x \ge -5$. These are redundant. It is enough to simply say $x \ge \frac 12$.
Now to do it.
$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$
$(3x+4) + (2x-1) -\frac 14(x + 5) = 5$
$5x +3 - \frac x4 - \frac 54 = 5$
$\frac {19}{4}x + \frac 74 = 5$
$19x + 7 = 20$
$19x = 13$
$x = \frac {13}{19}$.
Now I must check does this agree with my assumption that $x \ge \frac 12$. Yes it does.
$x = \frac {13}{19}$ is one possible answer..
Case 2:
$|3x + 4| = 3x+4$ because $3x + 4 \ge 0$ and $|2x-1| = 2x -1$ because $2x -1 \ge 0$ and $|x+5| = -(x+5)$ because $x + 5 < 0$.
First check our assumptions. As above: $x \ge -\frac 43$ and $x\ge \frac 15$ !!BUT!!! $x + 5 < 0 \immplies x < -5$. This is a contradiction. Case 2 is impossible.
Case 3: $|3x + 4| = 3x + 4$ and $x \ge -\frac 43$ and $|2x -1| =-(2x-1)$ and $x < \frac 12$ and $|x+5|=x+5$ and $x \ge -5$. In other words $-\frac 43 \le x < \frac 12$
Then
$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$
$(3x+4) - (2x-1) -\frac 14(x + 5) = 5$
$x + 5 -\frac 14(x+5) = 5$
$\frac 34(x+5) = 5$
$x + 5 = \frac {20}3$
$x = \frac 53$
But this contradicts $x < \frac 12$. So Case 3: is impossible.
Case 4: $|3x+4|=3x +4$ so $x \ge -\frac43$ and $2x-1 =-(2x-1)$ so $x < \frac 12$ and $|x+5|=-(x+5)$ so $x < -5$. This is impossible.
Case 5 & 6: $|3x+4| = -(3x + 4)$ so $x < -\frac43$ and $2x-1=2x + 1$ so $x \ge 12$. That's impossible.
Case 7: $|3x+4| = -(x+4)$ so $x < -\frac 43$ and $2x-1=-(2x +1)$ so $x < \frac 12$ and $|x+5|= x+5$ so $x \ge -5$. In other words $-5 \le x <-\frac 34$
Then
$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$
$-(3x+4) - (2x-1) -\frac 14(x + 5) = 5$
$-5x - 3 -\frac 14 x -\frac 14 5= -\frac {21x}4 - \frac {17}4 = 5$
$-21x - 17 = 20$
$-21x = 3$
$x =-\frac 3{21}$. But this contradicts $x < -\frac 43$
Case 8: $|3x+4| = -(x+4)$ so $x < -\frac 43$ and $2x-1=-(2x +1)$ so $x < \frac 12$ and $|x+5|= -(x+5)$ so $x < -5$. In other words $x < -5$.
Then
$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$
$-(3x+4) - (2x-1) +\frac 14(x + 5) = 5$
$-\frac {19}{4}x - \frac 74 = 5$
$-19x - 7 = 20$
$-19x = 27$
$x = -\frac {27}{19}$
This contradicticts $x < -5$.
So the ONLY solution is: $\frac {13}{19}$
Now notice half those cases were contradictory from the get-go.
Had we been efficient we would have noticed that:
$|3x +4| = \pm (3x+4)$ or $x \ge;< \frac 43$ and $|2x -1| = \pm(2x + 1)$ or $x \ge;< \frac 12$ and $|x+5|=\pm (x+5)$ or $x \ge;< -5$ center around three DEPENDENT critical points: $\frac 12, -\frac 43, -5 $ and these 8 cases break down to an actual four cases:
Case A: $x \ge \frac 12$
Case B: $-\frac 43 \le x < \frac 12$
Case C: $-5\le x < -\frac 43$
Dase D: $-5 < x$.
In the range $x<-5$ (note that when $x=-5, x+5=0$), we have $$|neg|+|neg|-\frac14|neg|=5$$ as all three expressions within the mod signs are negative in that range.
Thus we can solve: $$-(3x+4)-(2x-1)+\frac14(x+5)=5$$ and see that $x=-\frac{27}{19}\not<-5$ and therefore in this range there are no solutions.
In the range $-5<x<-\frac43$ (note that when $x=-\frac43, 3x+4=0$), we have $$|neg|+|neg|-\frac14|pos|=5$$ So we solve in the range $-5<x<-\frac43$: $$-(3x+4)-(2x-1)-\frac14(x+5)=5$$ and achieve $x=-\frac{37}{21}$ which IS in the correct range and therefore a solution.
Can you continue this?