Solve $|x-1|-2|x+5|>3+x$.
Problem
Source: Kosovo MO 2014 Grade 9, Problem 2
Tags: algebra
franzliszt
07.06.2021 21:38
Not always true. Take $x=1$ for example.
juliankuang
07.06.2021 22:01
ze solution shalt be;
We shalt casework.
If $-5 < x \le 1$, We shalt have $1 - x - (2x + 10) > 3 + x$. Thus It must be that $-12 > 4x,$ and thus it is that $x < -3$.
If it turns out that $x > 1$, We must get $x - 1 - 2x - 10 > 3 + x$, thus $-x - 11 > 3 + x$, so we have $x < -7$. As it is defined that $x > 1$, this ain't possible boi.
If $x \le -5$, We are to have $x + 11 > 3 + x$. Try to make this false, you won't so don't try actually.
Combining the intervals, we have $\boxed{x < -3}$.
pinkpig
07.06.2021 22:05
There are three cases to this question. First, assume that $x>1.$ Therefore, the equation simplifies to $x-1-2(x+5)>3+x$ which simplifies to $-x-11>3+x.$ This implies that $-7>x$ which leads to no solutions for case one. Second, assume that $-5\leq(x)\leq1.$ Therefore, the equation simplifies to $1-x-2(x+5)>3+x$ which simplifies to $-9-3x>3+x.$ This implies that $-3>x.$ Therefore, the solutions to this case will be $-5\leq(x)<3$ Lastly, assume that $x<-5.$ Therefore, the equation simplifies to $1-x+2(x+5)>3+x$ which simplifies to $x+11>3+x.$ This is true for all x so the solution to this case is $x<-5$ Finally, we add up all the cases to conclude that the final answer is $x<-3$ Hope this helps! Please tell me if there are any errors in my solution.
juliankuang
07.06.2021 22:13
Yup! I have the same thing, this looks good <3