Question

I-x+4I<=1 solve for x

Answer 1

\[-x+4\le 1\]and\[x-4\le 1\]

Answer 2

when (-x+4)>=0 ,x<=4
|-x+4|=-x+4<=1
x>=3
so 3<=x<=4(result 1)

when (-x+4)<0, x>4
|-x+4|=-(-x+4)=x-4<=1
x<=5
so 4<x<=5(result 2)

according to result1 and result

3<=x<5

Answer 3

@JohnHanShanghai is right !

Answer 4

I-x+4I<=1

-x+4 <= 1 (or) -x+4 >= -1
-x <= -3 (or) -x > = -5
x >= 3 (or) x <= 5

3<= x <= 5

note that 5 is also included in the solution

Answer 5

i hate \(-x\) so i would start with the fact that \(|-x+4|=|x-4| \) i.e. start with
\[|x-4|\leq 1\]
\[-1\leq x-4\leq 1\]
\[3\leq x\leq 5\]
just a matter of taste is all

Answer 6

keeps you from having to flip the inequality

Answer 7

i hate -x too :)

Answer 8

|-x+4|≤1
√((-x+4)^2 ) ≤1
〖(-x+4)〗^2 ≤1^2
Expanding LHS:
x^2-8x+16≤1
x^2-8x+15≤0
Factorising the LHS gives
(x-3)(x-5)≤0
The solution is therefore 3≤x≤5.that is values from 3 right up to 5.