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Question 1 of 5
Illustrate |x|=3 using a number line.
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Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies.
First, find the possible values of x
|x|=3 means the distance of x from 0 is 3 units. Therefore, x=3 or x=-3
Next, plot 3 and -3 on the number line.
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Question 2 of 5
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Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies.
Since we are finding the absolute value of x, form a positive and negative equation and solve for x on both equations
Positive:
x-5 |
= |
11 |
x-5 +5 |
= |
11 +5 |
Add 5 to both sides |
x |
= |
16 |
Negative:
x-5 |
= |
-11 |
x-5 +5 |
= |
-11 +5 |
Add 5 to both sides |
x |
= |
-6 |
To check if the values of x are correct, substitute the values to the equation and check if its absolute value is 11
x=16
|x-5| |
= |
11 |
|16-5| |
= |
11 |
x=16 |
|11| |
= |
11 |
11 |
= |
11 |
The absolute value of 11 is 11 |
x=-6
|x-5| |
= |
11 |
|-6-5| |
= |
11 |
x=16 |
|-11| |
= |
11 |
11 |
= |
11 |
The absolute value of -11 is 11 |
Therefore, the absolute values of x is 16 or -6
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Question 3 of 5
Solve for x
4|2x-2|-14=10
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First, convert the equation to its standard form
4|2x-2|-14 |
= |
10 |
4|2x-2|-14 +14 |
= |
10 +14 |
Add 14 to both sides |
4|2x-2| |
= |
24 |
4|2x-2| ÷4 |
= |
24 ÷4 |
Divide both sides by 4 |
|2x-2| |
= |
6 |
Since we are solving an absolute value equation, form a positive and negative equation and solve for x on both equations
Positive:
2x-2 |
= |
6 |
2x-2 +2 |
= |
6 +2 |
Add 2 to both sides |
2x |
= |
8 |
2x ÷2 |
= |
8 ÷2 |
Divide both sides by 2 |
x |
= |
4 |
Negative:
2x-2 |
= |
-6 |
2x-2 +2 |
= |
-6 +2 |
Add 2 to both sides |
2x |
= |
-4 |
2x ÷2 |
= |
-4 ÷2 |
Divide both sides by 2 |
x |
= |
-2 |
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Question 4 of 5
Solve for x
-4|1-34x|=-16
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First, convert the equation to its standard form
-4|1-34x| |
= |
-16 |
|
-4|1-34x| ÷(-4) |
= |
-16 ÷(-4) |
Divide both sides by -4 |
|
|1-34x| |
= |
4 |
Since we are solving an absolute value equation, form a positive and negative equation and solve for x on both equations
Positive:
1-34x |
= |
4 |
|
1-34x -1 |
= |
4 -1 |
Subtract 1 from both sides |
|
-34x |
= |
3 |
|
-34x ×(-4) |
= |
3 ×(-4) |
Multiply both sides by -4 |
|
3x |
= |
-12 |
3x ÷3 |
= |
-12 ÷3 |
Divide both sides by 3 |
x |
= |
-4 |
Negative:
1-34x |
= |
-4 |
|
1-34x -1 |
= |
-4 -1 |
Subtract 1 from both sides |
|
-34x |
= |
-5 |
|
-34x ×(-4) |
= |
-5 ×(-4) |
Multiply both sides by -4 |
|
3x |
= |
20 |
3x ÷3 |
= |
20 ÷3 |
Divide both sides by 3 |
|
x |
= |
203 |
x |
= |
623 |
Convert to a mixed number |
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Question 5 of 5
Solve for y
-7|2y-2|+5=-37
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First, convert the equation to its standard form
-7|2y-2|+5 |
= |
-37 |
-7|2y-2|+5 -5 |
= |
-37 -5 |
Subtract 5 from both sides |
-7|2y-2| |
= |
-42 |
-7|2y-2| ÷(-7) |
= |
-42 ÷(-7) |
Divide both sides by -7 |
|2y-2| |
= |
6 |
Since we are solving an absolute value equation, form a positive and negative equation and solve for x on both equations
Positive:
2y-2 |
= |
6 |
2y-2 +2 |
= |
6 +2 |
Add 2 to both sides |
2y |
= |
8 |
2y ÷2 |
= |
8 ÷2 |
Divide both sides by 2 |
y |
= |
4 |
Negative:
2y-2 |
= |
-6 |
2y-2 +2 |
= |
-6 +2 |
Add 2 to both sides |
2y |
= |
-4 |
2y ÷2 |
= |
-4 ÷2 |
Divide both sides by 2 |
y |
= |
-2 |
To check if the values of y are correct, substitute the values to the equation and check if its absolute value is -37
y=4
-7|2y-2|+5 |
= |
-37 |
-7|2(4)-2|+5 |
= |
-37 |
y=4 |
-7|8-2|+5 |
= |
-37 |
-7|6|+5 |
= |
-37 |
-7(6)+5 |
= |
-37 |
The absolute value of 6 is 6 |
-42+5 |
= |
-37 |
-37 |
= |
-37 |
y=-2
-7|2y-2|+5 |
= |
-37 |
-7|2(-2)-2|+5 |
= |
-37 |
y=-2 |
-7|-4-2|+5 |
= |
-37 |
-7|-6|+5 |
= |
-37 |
-7(6)+5 |
= |
-37 |
The absolute value of -6 is 6 |
-42+5 |
= |
-37 |
-37 |
= |
-37 |
Therefore, the absolute values of y is 4 or -2