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Question 1 of 4
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Terms with the same variable can be added or subtracted.
Both terms have a as their variable, meaning we can freely add or subtract them like integers.
Since the signs are the same (- -) we will find the sum of the two numbers
8+5 |
= |
13 |
Find the Sum |
Sum |
= |
13 |
Insert the sign of the larger number (-8) to the difference as well as the variable a.
-8a-5a |
= |
-13a |
Insert a negative sign since the larger number, 8 is negative |
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Question 2 of 4
Evaluate
9x-(-3x)-16-(-7x)+24
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Terms with the same variable can be added or subtracted.
First, solve for the values with an x variable.
9x-(-3x)-(-7x) |
= |
9x+3x+7x |
Recall,  |
|
= |
19x |
Next, solve for the values with no variables.
Since the signs are the same (- +) we will find the difference of the two numbers
24-16 |
= |
8 |
Find the Difference |
Difference |
= |
8 |
Insert the sign of the larger number (+24) to the difference.
-16+24 |
= |
8 |
Use a positive sign since the larger number, 24 is positive |
Finally, combine the terms.
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Question 3 of 4
Evaluate
7x-(-x)-12-(+15x)+30
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Terms with the same variable can be added or subtracted.
First, solve for the values with an x variable.
7x-(-x)-(+15x) |
= |
7x+x-(+15x) |
Recall,  |
7x+x-(+15x) |
= |
7x+x-15x |
Recall,  |
|
= |
8x-15x |
Since the signs are different (+ -) we will find the difference of the two numbers
15-8 |
= |
7 |
Find the Difference |
Difference |
= |
7 |
Insert the sign of the larger number (-15) to the difference as well as the variable x.
8x-15x |
= |
-7x |
Insert a negative sign since the larger number, 15 is negative |
Next, solve for the values with no variables.
Since the signs are different (- +) we will find the difference of the two numbers
30-12 |
= |
18 |
Find the Difference |
Difference |
= |
18 |
Insert the sign of the larger number (+30) to the difference.
-12+30 |
= |
18 |
Use a positive sign since the larger number, 30 is positive |
Finally, combine the terms.
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Question 4 of 4
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Rules for Dividing Integers
Transform the expression into a fraction to make it easier to solve.
−32y3−8y |
= |
4y3y |
Recall,  |
|
|
= |
4y3-1 |
Recall rules for indices |
|
= |
4y2 |
|
= |
4y2 |