Solving Absolute Values W/Variables On Both Sides by Lucretia Glover

Homework 2.2 - Journal Writing w/Examples

lglover3 — 2016-10-09 02:58:36 UTC

Greetings students,

Your first task will be to review the attached video link on "solving absolute value equations with variables on both sides". Take notes as you watch the video. When done, click anywhere on the screen and add a brief summary of what you learned. To support your understanding create two additional examples of your own and solve them. When done, post your examples with the work within the same post.

To begin, click on the video below.

Viviana Barberi Homework

2016-10-09 22:47:30 UTC

Absolute value equations with variables on both sides. When the equation is in the positive case it is the same number so you set both sides equal to each other. When it is negative you multiply the side without the absolute value by -1. The equation is Iax+bI=cx+d. A positive case would look like ax+b=cx+d. A negative case would look like ax+b=-(cx+d). An extraneous solution is a solution to an equation that must be rejected because it doesn’t satisfy the original equation.There are cases with no extraneous solutions so always try two cases and check the answers by plugging them back into the original equation.

Example 1:
Ix-1I=2x+3
x-1=2x+3
-x=4
x=-4
-4-1=-5---5
-8+3=-5
This is an extraneous solution.
x-1=-2x-3
3x=-2
x=-2/3
5/3
5/3
X=-2/3 is the solution.

Example 2:
IX+2I=4x-1
x+2=4x-1
3=3x
x=1
3
3
x=1 is a solution.

x+2=-4x+1
5x=-1
x=-1/5
9/5
-9/5
This is an extraneous solution.

Linnea McWiliam Homework

lmcwilliam — 2016-10-11 18:11:16 UTC

Absolute value equations with variables on both sides. First the equation for the absolute value function with variables on both sides is |ax+b|=cx+d. They're three cases. Once case would be when the equation in the positive case is the same, set each side to equal each other. The positive equation would look like ax+b=cx+d because they are equal to each other. Another case would be when it is negative, multiply the side that does not have the absolute value, by -1. An example of a case that is negative the equation would be ax+b= -(cx+d), and with this equation you would distribute the negative to the cx+d. The definition of an extraneous solution that is discussed in the video, is an apparent solution to an equation that must be rejected because it does not satisfy the original equation. There are two different kinds of cases in order to find the solutions. There are also cases where there are no extraneous solutions. Plug in the solution to the original equation
Example:
|x-7|=2x+2
-x=9
x=-9
x-7=-2x-2
3x=5
x=5/3

Example:
|x+4|=3x-2
x+4=3x-2
-2x=-6
x=3

x+4=-3x+2
4x=-2
x=-2/4

Sophie Watkinson Homework

2016-10-11 19:48:00 UTC

I learned how to solve absolute value equations with variables on both sides and check if the answers I got were extraneous or not. An absolute value equation with variables on both sides looks like |ax+b|=cx+d. When solving them, you have to do it in two equations: one assuming the absolute value is positive and one assuming it is negative. The positive one looks like this: ax+b=cx+d. The negative assumption is set up like this: ax+b= -(cx+d). Then you solve for them in two different "cases" ,as the video calls them. After you are done solving both cases, you then have to check for an extraneous solution. Extraneous solutions are solutions that are apparent solutions to an absolute value equation that is not actually the solution to it. Checking for extraneous solutions are pretty simple, all you have to do is plug in each values of x into the original equation, and then solve. If you end up with the solution |x|=x then you know that the value of x you got is the correct the solution and not an extraneous one. If |x| does not equal x then the solution is extraneous, and the value of x you got was not correct.

Example 1:
|x+8|=2x-4
c1: 12=x is a solution
c2: x+8=-2x+4
x=-4/3 is an extraneous solution

Example 2:
|x+10|=10x-8
c1: 2=x is a solution
c2: x+10=-10x+8
-2/11=x is an extraneous solution.

Valerie Blinder Homework

vblinder — 2016-10-11 22:59:08 UTC

In the video, I learned how to solve absolute value equations where there are variables on either side of the equation sign (|ax + b| = cx + d). The video began by clarifying that, when a number is inside absolute value bars, the number becomes positive. I learned two different cases that are necessary to use when solving these absolute value equations. The first case (|x| = x, if x > 0) is ax + b = cx + d. The second case (|x| = -x, if x < 0) is ax + b = - (cx + d). When solving for an equation, you must solve it using the two cases, which results in two apparent solutions. Usually, with apparent solutions, one is the solution while the other is extraneous solution (although it is possible not to get an extraneous solution). An extraneous solution is an apparent solution to an equation that must be rejected because it does not satisfy the original equation. To sum up what I learned in the video, when solving for absolute value equations with two variables on either side of the equation, there are a few steps to follow. First, you should consider the two different cases. Then, after finding the two apparent solutions, substitute them for x in the original equation. One or both of these solutions will work in the original equation. Also, you should keep in mind that, just because you got a fraction as the extraneous solution for one problem, it doesn't mean that a fraction will always be the extraneous solution. Below are my two examples:

Example One:
|4x - 3| = 2x +1
C1: 4x - 3 = 2x + 1
2x - 3 = 1
2x = 4
x = 2 (This is the solution)
C2: 4x - 3 = - (2x + 1)
4x - 3 = -2x - 1
6x - 3 = -1
6x = 2
x = 2/3 (This is
extraneous)

Example Two:
|x - 1| = 3x
C1: x - 1 = 3x
x = 3x +1
-2x = 1
x = - 1/2 (This is
extraneous)
C2: x - 1 = - (3x)
x - 1 = -3x
x = -3x + 1
4x = 1
x = 1/4 (This is the
solution)

Eve Butler Homework

ebutler9 — 2016-10-12 19:32:40 UTC

In this video, I learned to solve absolute value equations when there are variables on each side of the equation sign. The video began by explaining that when you have an equation as such, you need to find the answer but solving the equation in two different cases. Case Number One states that the equation in absolute value in equal to the one on the other side of the equation. The equation is set up like this: ax+b=cx+d. To solve this, you keep the equation the same and solve as you would any other problem. The second case changes the cx+d into it's opposite. You solve like this: ax+b = -(cx+d). After solving the equation in both cases, you substitute your answer back in to find the answer. Sometimes, both cases can satisfy the solution, but not always. The extraneous solution is the apparent solution that doesn't satisfy the equation and is rejected.

Example 1:
|x-2| = 2x+5

Case 1:
x-2 = 2x+5
-x -x
-2 = x+5
-5
x = -7 (SOLUTION)

Case 2:
x-2 = -(2x+5)
x-2 = -2x-5
+2x
3x-2 = -5
+2
3x = -3
x = -1 (ALSO A SOLUTION)

Example 2:
|x-5| = 2x -6

Case 1:
x-5 = 2x -6
-x -x
-5 = x -6
+6
x = 1 (SOLUTION)

Case 2:
x-5 = -(2x -6)
x-5 = -2x +6
+2x
3x-5=6
+5
3x = 11
/3 /3
x = 11/3 (ALSO A SOLUTION)

Jessica Mann Homework

2016-10-12 20:52:44 UTC

In this video, I learned about how to properly solve absolute value functions that have variables on both sides. I also learned how to look for extraneous solutions. Extraneous solutions are solutions that are not actually able to make the function true. To solve for equations with variables on both sides, you use the two formulas, ax+b=cx+d and ax+b=-(cx+d). Many times both of these equations will result in sufficient solutions, but you must always check to see if either of the equations gives an extraneous answer.

Example 1:
|x-3|=3x+2
Case 1:
x-3=3x+2
-x -x
-3=2x+2
-2 -2
-5=2x
x= -2.5 (Solution)
Case 2:
x-3=-3x-2
-x +2 -x+2
-1=-4x
x=-1/4 ( not a solution)

Example 2:
|x+1|=2x-2
Case 1:
x+1=2x-2
-x -x
1=x-2
+2 +2
x=3 (solution)

Case 2:
x+1=-2x+2
+2x +2x
3x+1=2
-1 -1
3x=1
x=1/3 (not a solution)

Lara Thain Homework

lthain — 2016-10-12 21:36:36 UTC

In this video, I leant how to solve absolute value equations, with variables on both sides. To do this you have to consider two cases. If the variable is negative, or positive. For example, you might have an equation that looks like this: |x-4|=2x+2. Using this equation, for the positive side, you get x-4=2x+2, and the negative side, x-4=-(2x+2). This is why you have a extraneous solution as well as a regular solution. Once you do both , you plug them into the equation for each case, and see which of them was the solution and which was extraneous.

Example 1:
|x-4|=2x+2

Case 1:
x-4=2x+2
x=-6
-6-4=-12=2
-10=-10 (Solution)

Case 2:
x-4=-(2x+2)
x=-2
-6=-4+2
-6=-2(extraneous solution)

Example 2:
|x-1|=2x+2

Case 1:
x-1=2x+2
x=-3
-3-1=-6+2
-4=-4 (solution)

Case 2:
x-1=-(2x+2)
x=1/3
2/6-6/6=-4/6-4/6
-4/6=0 (extraneous solution)

Maddy Barket Homework

2016-10-18 21:32:04 UTC

In this video, I learned how to solve absolute value equation with variables on both sides. I learned that when solving an absolute value function, you have to be considerate of the two different cases. A way to think of this can be that one case is negative, and the other is positive.

Example 1: |2+v|=2

Case 1 (positive):
2+v=2
-2 -2
v= 0

Case 2:
2+v=-2
-2 -2
v=-4

I also learned what an extraneous solution is. An extraneous solution is an apparent solution that is rejected because it does not satisfy the equation.

Example 2:
|x+2|=2x-5

Case 1:
x-2=2x-5
-x -3
x=3

Case 2:
x-2= (2x-5)
x-2= -2x+5
3x=7
x=7/3

If you were to check these solution by plugging x back in, you will see that case 2 will say that x=1/3. This is an example of an extraneous equation because the solution does not satisfy the equation.

Sabrina Gordon Homewor

ssgordon1 — 2016-10-18 23:10:58 UTC

In this video, I learned how to solve absolute value equations with variables on both sides of the equation sign. An example of this would be :
|ax+b|=cx+d
To solve for this, you must consider the two cases, the negative and the positive one. An example of this would be: |x-3|=2x+4. Using this equation for the positive side, you get x-3=2x+4, and for the negative side you get, x-3=-(2x+4). This is why you have an extraneous solution as well as a regular solution. Once both of them are put into the positive and negative forms, you plug them into the equation for each case, and see which of them has a solution and which was extraneous. An example of this would be:
Example 1:
|x-3|=2x+4
Case 1:
x-3=2x+4
-x -x
-3=x+4
-4 -4
-7=x (solution)
Case 2:
x-3=-2x-4
-x -x
-3=-3x-4
+4 +4
1=-3x
/-3 /-3
1/-3 (extraneous solution)

Christina Zahringer HW

2016-10-19 02:40:08 UTC

During this video, I learned how to solve absolute value equations when there are two variables; one on each side of the equation. The video explained that when you find an equation like this, you need to have two cases; case 1 and case 2. This is how the equation is set up:
|ax+b| = cx +d. For case 1, you need to assume that the value of the variable inside the absolute value brackets is positive, and equal to whatever is on the other side of the equation. After solving for case 1, you need to solve for case 2. In case 2, you need to assume that what is inside the absolute value brackets is negative, and is not equal to what is on the other side of the equation. While watching the video I learned what an extraneous solution is. An extraneous solution is a solution to an equation that must be rejected because it doesn't satisfy the original equation. Lets say you are given the equation: |2x+12| = 4x
To solve for Case 1:
2x+12=4x
-2x -2x
12=2x
x = 6 (solution)

To solve for Case 2:
2x+12= -4x
-2x -2x
12 = -6x
x = -2 (solution)
After solving for Case 1 and Case 2, you need to plug your answers back into the original equation to check your work.
Checking:
2(6) +12 = 4(6)
12 +12 = 24
24 = 24 -- This checks out
Now you have to plug in for the other value:
2(-2)+12= 4(-2)
-4+12=-8
+4 +4
12 = 12 -- This also checks out

Example 2:
|4x-3| = 2x +1
To solve for Case 1:
4x-3=2x+1
+3 +3
4x= 2x +4
-2x -2x
2x = 4
X = 2 (solution)

To solve for Case 2:
4x-3=-2x-1
+3 +3
4x = -2x +2
+2x +2x
6x=2
x = 2/3 (extraneous solution)

Plug in for Case 1:
4(2)-3=2(2)+1
8-3=4+1
5 = 5 -- This checks out

Plug in for Case 2:
4(2/3)-3=2(2/3)+1
8/3-3/1= 4/3+1
-1/3=7/3 -- This doesn't check out