# How do I solve this problem? 6-|3-y|=1I'm 12 so simple answers would be good. I'm in algebra 1.

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### 8 Answers

*6-|3-y|=1*

*Let's look at it with some simple answers. The absolute value part of it is what's giving you trouble, I'm sure, so put you finger over that part of it and look at the question this way:*

*Six minus *what *equals one?*

When you do that, it's easy to see that *|3-y| is 5!*

*Here's the tricky part. The easy way to remember what absolute value means is "the distance from zero on the number line." So if something has an absolute value of 5, it's five away from zero, which means it's either 5 or -5.*

*So there will usually be two answers to an absolute value problem. In this case, 3-y can be either 5 or -5. So if you solve these two equations:*

*3-y = 5*

*3-y = -5*

You'll have the answer!

*6-|3-y|=1*

Notice the lines on both sides of 3-y, that is called absolute value which means positive value, first we need to remove these lines, in order to remove them negative signs must be converted to positive signs

6 - |3 - y| =1

6 - 3 + y =1

Now we need to isolate y, to find it's value,

3 + y =1

3 + y - 3 = 1 - 3 (subtract 3 from both sides)

**y = -2 Answer**

*6-|3-y|=1*

6 - 3 + y = 1

combine terms

3 + y = 1

subtract the y

y = -2

*6-|3-y|=1*

6 - 3 + y = 1

combine terms

3 + y = 1

subtract the y

y = -2

6 - 3 + y = 1

3 + y = 1

y = 1 - 3

y = - 2

The negative before the parenthasis makes the 3 a negative 3 and the y a positive y. Then when you do 6 - 3, you get 3. If you subtract the 3 to get it on the other side, you get 1 - 3. That leaves you with y = - 2.

Absolute in maths stands for the positive value . We enclose the numeric or algbraic expression between two vertical lines to mean the absolute value of the expressio. Examples:

|-100|=100, |-5.78|= 5.78.

|6|=6, |3.2|=3.2.

|x|= -x, if x is negative and |x|= X if x is positive.

Also see how they mean here:

If f(x) is a function of x, then |f(x)|<=10 means -10<=f(x)<=10.

x^2-2 is greater than 2 means, x^2-2>=2

x^2>=4 or |x|>=2 or x>2 or x <-2.

5

y4(y5)215

y7(y2

-3y+6=3(y-2)