Intermediate Algebra Tutorial 7


Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 7: Linear Equations in One Variable


WTAMU > Virtual Math Lab > Intermediate Algebra > Tutorial 7: Linear Equations in One Variable


 

checkAnswer/Discussion to 1a

problem 1a
 

ad1a

*Inverse of add. 5 is sub 5 from both sides
 

*Inverse of mult. by 6 is div. both sides by 6

 
 

If you put 0 back in for y in the original problem you will see that 0 is the solution we are looking for.

(return to problem 1a)
 


 

checkAnswer/Discussion to 1b

problem 1b
 

ad1b

*Get all x terms on one side
 

*Inverse of sub. 3 is add. 3 to both sides
 

*Inverse of mult. by 2 is div. both sides by 2
 
 

If you put 5 back in for x in the original problem you will see that 5 is the solution we are looking for.

(return to problem 1b)


 

checkAnswer/Discussion to 1c

problem 1c
 

ad1c

*Distribute the inner most (   )
*Distribute the outer grouping [  ]
 

*Get all t terms on one side
 

*Inverse of mult. by 11 is div. both sides by 11
 

 
 

If you put 8/11 back in for t in the original problem you will see that 8/11 is the solution we are looking for.

(return to problem 1c)
 


 

checkAnswer/Discussion to 1d

problem 1d
 

ad1d
*Mult. both sides by the LCD of 20
 

*Use dist. property to remove (  )

*Get all x terms on one side

*Inverse of add. 108 is sub. 108
 

*Inverse of mult. by -51 is div.  by -51
 
 

If you put 3 back in for x in the original problem you will see that 3 is the solution we are looking for.

(return to problem 1d)
 


 

checkAnswer/Discussion to 1e

7(x - 3) = 7x + 2
 

ad1e
*Use dist. property to remove (  )

 *Get all x terms on one side
 

Since the variable dropped out AND we ended up with a false statement, the answer is no solution.

(return to problem 1e)
 

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WTAMU > Virtual Math Lab >Intermediate Algebra >Tutorial 7: Linear Equations in One Variable


Last revised on July 1, 2011 by Kim Seward.
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