+0

+1
13
1
+273

What values of x satisfy $$|x-4|+2|x+3|\le11$$? Express your answer in interval notation.

Mar 30, 2024

#1
+467
+1

To solve the inequality $$|x - 4| + 2|x + 3| \leq 11$$, we'll consider different cases depending on the sign of $$x - 4$$ and $$x + 3$$.

Case 1: $$x - 4 \geq 0$$ and $$x + 3 \geq 0$$

In this case, both absolute values are positive, so we have:

$(x - 4) + 2(x + 3) \leq 11$

$x - 4 + 2x + 6 \leq 11$

$3x + 2 \leq 11$

$3x \leq 9$

$x \leq 3$

Case 2: $$x - 4 \geq 0$$ and $$x + 3 < 0$$

In this case, $$x - 4 \geq 0$$ and $$x + 3 < 0$$, so $$x$$ is between -3 and 4.

$(x - 4) + 2(-x - 3) \leq 11$

$x - 4 - 2x - 6 \leq 11$

$-x - 10 \leq 11$

$-x \leq 21$

$x \geq -21$

Case 3: $$x - 4 < 0$$ and $$x + 3 \geq 0$$

In this case, $$x - 4 < 0$$ and $$x + 3 \geq 0$$, so $$x$$ is between -3 and 4.

$-(x - 4) + 2(x + 3) \leq 11$

$-x + 4 + 2x + 6 \leq 11$

$x + 10 \leq 11$

$x \leq 1$

Case 4: $$x - 4 < 0$$ and $$x + 3 < 0$$

In this case, both absolute values are negative, so we have:

$-(x - 4) + 2(-x - 3) \leq 11$

$-x + 4 - 2x - 6 \leq 11$

$-3x - 2 \leq 11$

$-3x \leq 13$

$x \geq -\frac{13}{3}$

Now, let's combine the solutions from each case:

- Case 1: $$x \leq 3$$

- Case 2: $$x \geq -21$$

- Case 3: $$x \leq 1$$

- Case 4: $$x \geq -\frac{13}{3}$$

The overlapping intervals are $$[-21, 3]$$ and $$\left[ -\frac{13}{3}, 1 \right]$$.

Therefore, the values of $$x$$ satisfying the inequality are $$\boxed{[-21, 3] \cup \left[ -\frac{13}{3}, 1 \right]}$$.

Mar 30, 2024