Question

|х - 1| + |x+2|≤ |x-3|+4

Answer 1

$|x-1|+|x+2|\leq|x-3|+4$

Найдём нули каждого модуля и проверим, являются ли эти значения решениями неравенства:

$x-1=0;\ \ \ x=1\\|1-1|+|1+2|\leq|1-3|+4\\0+3\leq2+4\\3\leq6$

Верное неравенство. Число 1 является решением неравенства.  

$x+2=0;\ \ \ x=-2\\|-2-1|+|-2+2|\leq|-2-3|+4\\3+0\leq5+4\\3\leq9$

Верное неравенство. Число (-2) является решением неравенства.

$x-3=0;\ \ \ x=3\\|3-1|+|3+2|\leq|3-3|+4\\2+5\leq0+4\\7\leq4$

Неверное неравенство. Число 3 не является решением неравенства.

Полученные три точки разбивают числовую прямую на 4 интервала. В каждом интервале открываем модуль по правилу:

$|a|=\displaystyle\left \{ {{a;\ a\geq0} \atop {-a;\ a$

Открыв модули, решаем обычные линейные неравенства.

$1)\ x$  -  внутри каждого модуля будут отрицательные числа.

$-(x-1)-(x+2)\leq-(x-3)+4\\-x+1-x-2\leq-x+3+4\\-x-x+x\leq3+4-1+2\\-x\leq8;\ \ \ x\geq-8\\\\\displaystyle\left \{ {{x$

$2)~-2< x$  -  внутри второго модуля будет положительное число, внутри первого и третьего модуля - отрицательные числа.

$-(x-1)+(x+2)\leq-(x-3)+4\\-x+1+x+2\leq-x+3+4\\-x+x+x\leq3+4-1-2\\x\leq4\\\\\displaystyle\left \{ {{-2< x$

$3)~1< x$  -  внутри третьего модуля будет отрицательное число, внутри первого и второго модуля - положительные числа.

$(x-1)+(x+2)\leq-(x-3)+4\\x-1+x+2\leq-x+3+4\\x+x+x\leq3+4+1-2\\3x\leq6;\ \ \ \ x\leq 2\\\\\displaystyle\left \{ {{1$

$4)~x>3$   -  внутри трёх модулей будут положительные числа.

$x-1+x+2\leq x-3+4\\x+x-x\leq-3+4+1-2\\x\leq0\\\\\displaystyle\left \{ {{x>3} \atop {x\leq0}} \right. \ \ \ \Rightarrow\ \ \ \boxed{\boldsymbol{x\in\varnothing}}$

Объединяем все решения:

$[-8;-2)\cup\{-2\}\cup(-2;1)\cup\{1\}\cup(1;2]=[-8;2]$

Ответ: $x\in[-8;2]$

Answer 2

To solve the inequality |x - 1| + |x + 2| ≤ |x - 3| + 4, you can consider different cases based on the signs of the expressions inside the absolute values.

1. Case 1: x is less than 1 (x < 1)

In this case, both |x - 1| and |x + 2| become -(x - 1) and -(x + 2), respectively, as they are both negative. Similarly, |x - 3| becomes -(x - 3).

So, the inequality becomes:

-(x - 1) - (-(x + 2)) ≤ -(x - 3) + 4

Simplify it:

-(x - 1) + (x + 2) ≤ -(x - 3) + 4

Now, solve for x:

-x + 1 + x + 2 ≤ -x + 3 + 4

3 ≤ 7

This is always true for any x < 1. So, x < 1 is a valid solution in this case.

2. Case 2: 1 ≤ x ≤ 3 (1 ≤ x ≤ 3)

In this case, |x - 1| becomes (x - 1), |x + 2| remains the same, and |x - 3| becomes (3 - x).

The inequality becomes:

(x - 1) + |x + 2| ≤ (3 - x) + 4

Now, simplify it:

x - 1 + |x + 2| ≤ 7 - x

Now, consider two subcases for |x + 2|:

Subcase 2a: x + 2 is non-negative (x ≥ -2)

In this case, |x + 2| = (x + 2), and the inequality is:

x - 1 + (x + 2) ≤ 7 - x

Now, solve for x:

2x + 1 ≤ 7 - x

Add x and subtract 1 from both sides:

3x ≤ 6

x ≤ 2

Subcase 2b: x + 2 is negative (x < -2)

In this case, |x + 2| = -(x + 2), and the inequality is:

x - 1 - (x + 2) ≤ 7 - x

Now, simplify it:

-3 ≤ 7 - x

Add x and subtract 7 from both sides:

-x ≤ 4

Multiply by -1 (remember to reverse the inequality when multiplying by a negative number):

x ≥ -4

Now, considering that we are in the 1 ≤ x ≤ 3 range for Case 2, the valid solution is -4 ≤ x ≤ 2.

3. Case 3: x is greater than 3 (x > 3)

In this case, both |x - 1| and |x + 2| become (x - 1) and (x + 2), respectively. |x - 3| remains the same.

The inequality becomes:

(x - 1) + (x + 2) ≤ |x - 3| + 4

Now, simplify it:

2x + 1 ≤ (x - 3) + 4

2x + 1 ≤ x + 1

Subtract x and 1 from both sides:

x ≤ 0

So, the valid solution in this case is x ≤ 0.

Now, we have three different solution intervals:

1. x < 1
2. -4 ≤ x ≤ 2
3. x ≤ 0

To find the combined solution, you can express it as a union of these intervals:

x < 1 or (-4 ≤ x ≤ 2) or (x ≤ 0)

This is the final solution to the inequality |x - 1| + |x + 2| ≤ |x - 3| + 4.