Drug Rehab Centers in or near Westland, PA that accept the following payment types: Free or Low Cost

If you are looking for affordable drug and alcohol rehabilitation services in or near Westland, Pennsylvania, it might be a good idea to consider public insurance to help you access free or low cost addiction treatment.

You can get low cost and free insurance coverage if you meet certain eligibility requirements for programs such as Medicaid and Medicare. These programs are funded in partnership between state governments and the federal government with the goal of ensuring that more people are able to access the treatment services they need.

Medicare, for instance, is a program run by the federal government. It provides health care coverage for people above the age of 65 years as well as those who have certain disabilities.

This program is available to you regardless of your level of income. However, you will have to pay some premiums - although these are discounted to ensure that the insurance program is available.

On the other hand, Medicaid is a program run in partnership between the federal and state governments. It provides coverage for people who can prove that they have a low level of income.

Both Medicare and Medicaid can provide you with extensive coverage for substance abuse and addiction. However, you need to realize that the income qualifications will vary from one state to another. To determine if you are qualified for Medicaid, you should visit their official website. You can even sign up for coverage from this website.

While looking for affordable addiction treatment, you need to realize that not all drug rehab centers will accept Medicare and Medicaid. However, this is still a good option if you are looking for low cost treatment for your substance use disorder.

As long as you are qualified, these public health care insurance programs can help you access the free or low cost addiction treatment services that you need to overcome your substance abuse and addiction in or near Westland, Pennsylvania.