# How do you solve the inequality abs(x+3)>=abs(6x+9) and write your answer in interval notation?

Mar 29, 2017

-12/7≤x≤-6/5
$x \setminus \in \left[- \frac{12}{7} , - \frac{6}{5}\right]$

#### Explanation:

Solving the problem would be much easier if the absolute signs were removed. However, to do this, we need to divide the problem into cases depending on the signs of the expression inside. There are four possibilities:

Case 1: x+3≥0 and 6x+9≥0
Then the absolute signs don't really matter. We solve for x+3≥6x+9 to get 5x≤-6, or x≤-6/5. However, we said that x+3≥0 and 6x+9≥0. In other words, for this case, the solution has to be the intersection of x≤-6/5, x≥-3, and x≥-3/2. Thus, for this case, -3/2≤x≤-6/5.

Case 2: x+3≥0 and $6 x + 9 < 0$
Then, after removing the absolute signs, we must multiply the right-hand side by -1. We solve for x+3≥-6x-9 to get 7x≥-12, or x≥-12/7. However, we said that x+3≥0 and $6 x + 9 < 0$. In other words, for this case, the solution has to be the intersection of x≥-12/7, x≥-3, and $x < - \frac{3}{2}$. Thus, for this case, -12/7≤x<-3/2.

Case 3: $x + 3 < 0$ and 6x+9≥0
Then, after removing the absolute signs, we must multiply the left-hand side by -1. We solve for -x-3≥6x+9 to get 7x≤-12, or x≤-12/7. However, we said that $x + 3 < 0$ and 6x+9≥0. In other words, for this case, the solution has to be the intersection of x≤-12/7, $x < - 3$, and x≥-3/2. Thus, there are no solutions for this case.

Case 4: $x + 3 < 0$ and $6 x + 9 < 0$
Then, after removing the absolute signs, we must multiply both sides by -1. We solve for -x-3≥-6x-9 to get 5x≥-6, or x≥-6/5. However, we said that $x + 3 < 0$ and $6 x + 9 < 0$. In other words, for this case, the solution has to be the intersection of x≥-6/5, $x < - 3$, and $x < - \frac{3}{2}$. Thus, there are no solutions for this case.

We combine the solutions for the cases to get -12/7≤x≤-6/5. In interval notation, $x \setminus \in \left[- \frac{12}{7} , - \frac{6}{5}\right]$.