Jaison's Responses in WyzAnt Answers

Hi Karen,

Maybe this will help. Let's break it down into little pieces so bear with me! First, go by the DEFINITION of the absolute value. 

|x| = x if its positive or -x if x is negative. So, lets consider something simple. |x| = 3. You know that x = 3  and -3. So, you know that when dealing with absolute values, there will be 2 solutions instead of 1.

Now I'm not limited to using the absolute value on a single variable x. I can also take the absolute value of somehting more complicated like k - 4. Replace x with k - 4.

|k - 4| = k - 4 and -(k-4). So, I have two solutions just like my simple example above. The rest of the problem states that |k-4| > 1 right? Well, if since |k-4| = k-4 then k-4 > 1 AND |k-4| = -(k-4) then -(k-4) > 1. We have two inequalities because the absolute value has two solutions.

Now, let's solve for k in both equations. You get k > 5 and k < 3. k is just a label for the real number line. The two inequalities say that |k - 4| > 1 will be true if k > 5 and k < 3. You draw the number line and cover the numbers greater than 5 and the numbers < 3 with a thick line. Since the inequality is just greater than, you have to show that k = 5 and 3 are NOT part of the solution. You can do this by starting the thick lines at open circles that lie at k = 3 and 5.

It was long but I want you to understand how the absolute value works. Hope it helps!

Think of it as adding two numbers 16 + 1/2 (When you say "16 and a half", using the conjunction and is a way of saying add the two numbers) You can't add them until they have the same denominator. In front of 16 is the implied 1, or 16(1). It may seem silly to write (1) since multiplication by 1 does nothing but we can use it to find the least common denominator.

16(1) = 16 * (2/2) = (16 * 2) / 2 = 32 / 2

Now add the fractions

16 + 1/2 = 32/2 + 1/2 = 33/2

Plug in numbers for a, b, c. Let a = 2, b = 1, and c = 3 and see for yourself! When you are trying to determine the validity of a mathematical proposition, create a few concrete examples as a check. Use the simplest numbers whenever possible.

First, look at the terms, you have two instances of x and the rest of the terms are numbers. Since there aren't any other letters, this tells you that in this problem, x is an unknown, not a variable. You must isolate x on one side and that will give you the value of x.

Break up the problem into small pieces. When you see parenthesis and a number in front, that indicates that you must use the distribution property. There are two instances of this.

4(x - 1) + (-1)(x + 6) = 2

Distribute the 4

4x - 4 + (-1)(x + 6) = 2

Distrubte the -1

4x - 4 - x - 6 = 2

Now rearrange terms. You can do this because you can add a bunch of numbers in any order you want

4x - x - 4 - 6 = 2

Combine like terms, those with an x and those without.

3x - 10 = 2

Add 10 to both sides to remove it from the left side.

10 + 3x - 10 = 2 + 10

3x = 12

And finally, isolate x by dividing both sides of the equation by 3

x = 4

And there you have it! See, break it up into tiny pieces and the problem becomes much easier.

Hi Karlee,

Inequalities are a lot like regular equations except when there is an equal sign and you have x raised to the first power, there is one solution. You are going to find a set of numbers that will satisfy the inequality.

Since an inequality is an equation, we can use all of the same rules we use to move stuff from side to side of an equation with an equal sign with the exception of multiplying by -1. So, let's get x by itself.

First, divide through by 5 and move it to the right.

-x > 3

Second step: multiply through by -1. This reverses the inequality.

x < -3

x is a label for the real number line. This inequality says I want all x that are less than -3. On the real number line, this corresponds to all numbers to the left of -3. And since we do NOT have an equal sign, we CANNOT include -3 itself. 

So, we draw a line representing the real number line and we label a few numbers to the left and right of -3. We then draw a thick line starting at -3 extending to the left; this represents our solution. Now, we have to indicate that -3 is NOT part of the solution. One way of doing this is by putting an empty circle at -3 and starting the line at the circle.