Question

Please help me I've tried to learn this but it's just not working.
Solve the equation and check for extraneous solutions.
6|6-4x|=8x+4

Answer 1

Well, the first thing you want to do is completely isolate the absolute value part. SO this means dividing both sides by 6 first.

\[|6-4x| = \frac{ 8x+4 }{ 6 }\]
From here we write two equations. One of the equations we just simply drop the absolute value bars, which gives us this:

\[6-4x = \frac{ 8x+4 }{ 6 }\]
For the second equation, we must make the side opposite of the absolute value portion negative. Now always be careful when you do this, because its very easy to make a sign mistake. I suggest using parenthesis to make sure you do not make a mistake at this part. This is what you would have:
\[6-4x = -(\frac{ 8x+4 }{ 6 }) \]
So now we try to solve both equations. Starting with the first one, I can simplify that right side a little bit first, make the numbers smaller.
\[6-4x = \frac{ 4x }{ 3 }+\frac{ 2 }{ 3 }\]Now I split the fraction into 2 so you can maybe see what I did. Basically, I reduced the fraction by 2. Also, you are allowed to make a fraction for each term in the numerator. I had 2 terms, 8x and 4, so I made two fractions. It is not necessary, but I thought it might show you how I reduced it.

Okay, so now I want to get rid of those denominators on the right. Notice how both fractions have a 3 in the denominator. I can get rid of these by multiply everything by 3. Since the fractions are a division of 3, a multiplication of 3 will cancel them out. So this gives me:
\[3(6-4x) = \frac{ (3)4x }{ 3 }+\frac{ (3)2 }{ 3 } \]
\[18-12x = 4x + 2 \]
Now I can simply combine like terms and solve for x. So I will add 12x to both sides and I will subtract 2 from both sides. That will give me:
\[16 = 16x \implies x = 1\]
So this is a solution of the first equation. Now we can look at the second:

Now that negative I have in front I want to distribute inside of the parenthesis. When I do that, I get:
\[6-4x = \frac{ -8x-4 }{ 6 } \]
Now just like before, I can reduce this fraction also by 2. So that gives me:
\[6-4x = \frac{ -4x }{ 3 }-\frac{ 2 }{ 3 } \]
Once again. same situation as before, we just have different signs. And once again,. Ill multiply everything by 3 to get rid of those fractions on the left. That will leave me with:
\[18-12x = -4x - 2\]
So now I solve for x by adding 12x to both sides and adding 2 to both sides to give me:
\[20 = 8x \implies x = \frac{ 20 }{ 8 }= \frac{ 5 }{ 2 } \]
So those are the two answers we get. Now we actually have to check them, though. We could have done our math perfect, but its still possible these are not correct answers. We check by plugging them in to the original equation. So starting with x = 1
\[6|6-4(1)|=8(1)+4\]
\[6|2| = 12\implies 12 = 12\]
So that answer is true. Now let's check the second:
\[6|6-4(\frac{ 5 }{ 2 })| = 8(\frac{ 5 }{ 2 })+4\]
\[6|6-10| = 24 \implies 6|-4| = 24 \implies (6)(4) = 24 \implies 24 = 24\]So both of our answers are true, meaning we can say x = 1 and x = 5/2 are solutions.

Answer 2

@tryingtolovemath

Answer 3

omg! thank you so much! Was on the verge of a mental break down trying to figure this out T_T.

Answer 4

Hopefully what I put made sense :3

Answer 5

it did, it made perfect sense o.o

Answer 6

Lol, alright, awesome :3