Question 49225: 2=|4+3x| How do i solve problems like this?
|5x+1|<5 How do you isolate x when it is not an equation (<5 not =5)? Also,
what needs to be manipulated to graph this on a horizontal graph?
3+|x-3|=7 How do I solve this algebraically?
Thankyou.
Found 2 solutions by AnlytcPhil, tutorcecilia:
Answer by AnlytcPhil(1807)   (Show Source):
You can put this solution on YOUR website! 2=|4+3x| How do i solve problems like this?
|5x+1|<5 How do you isolate x when it is not an equation
(<5 not =5)? Also, what needs to be manipulated to graph this
on a horizontal graph?
3+|x-3|=7 How do I solve this algebraically?
Thankyou.
First here are the rules for getting rid of the absolute value symbols:
To remove absolute value symbols:
1. |expression| = A where A is a positive number write
two separate equations:
expression = A expression = -A
Solve each for the variable and there will usually be
two solutions.
2. |expression| < A where A is a positive number, write
this:
-A < expression < A
Then solve this for the variable in the middle.
The answer will usually be like this C < x < D. The graph
will be like this
-----------------(=======)-----------
C D
and in interval notation like this (C, D)
3. |expression| < A where A is a positive number, write
this:
-A < expression < A
Then you solve this for the variable in the middle.
The answer will usually be like this C < x < D. The graph
will be like this
-----------------[=======]-----------
C D
and in interval notation like this [C, D]
4. |expression| > A where A is a positive number becomes
expression < -A OR expression > A
Solve each for the variable with OR between them.
The answer will usually be like this "x < C OR x > D".
The graph will be like this
<=================)-------(=================>
C D
and in interval notation like this (-oo, C) U (D, oo)
5. |expression| > A where A is a positive number becomes
expression < -A OR expression > A
Solve each for the variable with OR between them.
The answer will usually be like this "x < C OR x > D".
The graph will be like this
<=================]-------[=================>
C D
and in interval notation like this (-oo, C] U [D, oo)
-----------------------------------------------
2 = |4+3x|
This is
|4+3x| = 2 or case 1 above. So write the two equations:
4+3x = 2 4+3x = -2
3x = -2 3x = -6
x = -2/3 x = -2
So the solution is these two values, -2/3 and -2
-------------------------------------------------
|5x+1| < 5
This is case 2: So write this
-5 < 5x+1 < 5
Add -1 to all three "sides":
-5 < 5x+1 < 5
-1 -1 -1
------------------
-6 < 5x < 4
Divide all three "sides" by 5
-6/5 < 5x/5 < 4/5
-6/5 < x < 4/5
The graph is
-----------------(========)-----------:
-6/5 4/5
or in interval notation: (-6/5, 4/5)
-----------------------------------------------
3 + |x-3| = 7
First get the absolute value term alone on the
left side by adding -3 to both sides, then you have
|x-3| = 4
This is case 1 above. So write the two equations:
x-3 = 4 x-3 = -4
x = 7 x = -1
So the solution is these two values, 7 and -1
Edwin
Answer by tutorcecilia(2152)  (Show Source):
You can put this solution on YOUR website! For absolute value problems, you will usually have two answers. Since we don't know if the value of "x" is negative or positive, both cases must be considered.
.
2=|4+3x|
Case # 1. [Just in case x = a positive number (x>0)]
2=4+3x [For case #1, just drop the absolute value bars and solve for x]
2-4=4-4+3x
-2=+3x
-2/3=+3x/3
(-2/3) = x
.
Case #2 [Just in case x = a negative number (x<0)]
2=|4+3x|
(-1)(2) = 4+3x [Drop the absolute bars, mulitiply one of the sides by (-1)]
-2 = 4+3x [Solve for x]
-2-4=4-4+3x
-6=3x
-2=x
.
x = (-2/3) or x = -2
.
Check by pluggin the values of "x" back into the original equation.
2=|4+3x|
2=4+3(-2/3)
2=4-2
2=2 [Checks out]
.
.
2=|4+3x|
-2=4+3(-2)
-2=4-6
-2=-2 [Also checks out, so both answers are correct]
.
.
|5x+1|<5 How do you isolate x when it is not an equation (<5 not =5)?
Solve this the same as above, except when multiplying or dividing by a negative number, switch the inequality sign.
|5x+1|<5 [Needs two cases in case x>0 or in case x<0]
.
Case #1. [Just in case x = a positive number (x>0)]
|5x+1|<5
5x+1<5 [Drop the absolute bars]
5x+1-1<5-1 [Solve for x]
5x<4
5x/5<4/5
x<4/5
.
Case #2.[Just in case x = a negative number (x<0)]
|5x+1|<5
5x+1<5 [Drop the absolute bars[
5x+1>(-1)5 [Multiply one side by (-1)and immediately switch the inequality sign]
5x+1>-5 [Solve for x]
5x+1-1>-5-1
5x>-6
5x/5>-6/5
x>-6/5
-6/5 or -1.2
.
x<4/5 AND x>-6/5 [Final answer]
[If the inequality is "less thAND" the graph will be "and". If the inequality is greatOR, the graph will be an "or".]
.
Also, what needs to be manipulated to graph this on a horizontal graph?
To graph you know that the two points will overlap because it is an AND graph (less thAND). Every point from -6/5 through 4/5 is included in the graph.
<______-2______(-1.2=======-1======0=====4/5) ___1______2_____>
.
The interval notation includes every point between (-1.2, 4.5)
.
.
3+|x-3|=7 How do I solve this algebraically?
.
Case #1.[Just in case x = a positive number (x>0)]
3+|x-3|=7
3+x-3=7 [Drop the absolute bars]
x=7 [Solve for x]
.
Case #2 [Just in case x = a negative number (x<0)]
3+|x-3|=7
3+x-3=7 [Drop the absolute bars]
3+x-3=(-1)(7) [Multiply one side by (-1)]
3+x-3=-7 [Solve for x]
x=-7
.
x = 7 or x = -7
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