Find the sum of values of $x$ that satisfies $\left | 3x-3 \right | + \left | 4x-4 \right | + \left | 5x-5 \right | \geq 24$
What I did so far:
$$\left | x-1 \right | \times \left | 3+4+5 \right | \geq 24$$
$$\left | x-1 \right | \times \left | 12 \right | \geq 24$$
And that is pretty much that because I never dealt with absolute values in multiplications before.
On
Factorise, $|x-1|$, so we have $$3|x-1|+4|x-1|+5|x-1|\geq24$$
Let $$|x-1|=t$$
So we have $$3t+4t+5t\geq24$$
or $$12t\geq24$$
or $$t\geq2$$
So $$|x-1|\geq2$$
There are two cases. If you want to know how to solve this comment here. The solution is:
$$x\leq-1$$ $$x\geq3$$
Still confused? Follow this: Step-by-Step Solution
The sum would (theoretically) be = -3
Notice that your linear term are factorable, they contain $|x-1|$, so we have $$3|x-1|+4|x-1|+5|x-1|\geq24$$ which results in $|x-1|\geq2$. it should be obvious that there are infinitely many $x$ values that work here. Not sure where the problem came from, but on a side note, had the inequality been "less or equal to 24" and $x$ been an integer, the problem would have made more sense.