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Beginning Algebra
Answer/Discussion to Practice Problems
Tutorial 14: Solving Linear Equations



 

checkAnswer/Discussion to 1a

problem 1a


 
ad1a
*Remove ( ) by using dist. prop.
 

*Inverse of add. 27 is sub. 27
 

*Inverse of mult. by -7 is div. by -7

 


 
If you put 4 back in for x in the original problem you will see that 4 is the solution we are looking for.

 

 


 

checkAnswer/Discussion to 1b

problem 1b


 
ad1b
*To get rid of the fractions, 
mult. both sides by the LCD of 8

*Get all the x terms on one side

*Inverse of sub. 1 is add. 1
 

 


 
If you put - 4 back in for x in the original problem you will see that - 4 is the solution we are looking for.

 

 


 

checkAnswer/Discussion to 1c

problem 1c


 
ad1c
*To get rid of the decimals,
mult. both sides by 100

*Get all the a terms on one side

*Inverse of sub. 5 is add. 5

*Inverse of mult. by 5 is div. by 5
 

 


 
If you put 4 back in for a in the original problem you will see that 4 is the solution we are looking for.

 

 


 

checkAnswer/Discussion to 1d

problem 1d


 
ad1d
*Remove ( ) by using dist. prop.

*Get all the x terms on one side
 


 
This time when our variable dropped out, we ended up with a FALSE statement.  Whenever that happens your answer is NO SOLUTION.

So, the answer is no solution.


 

 

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Last revised on July 26, 2011 by Kim Seward.
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