College Algebra Tutorial 14


College Algebra
Answer/Discussion to Practice Problems  
Tutorial 14: Linear Equations in One Variable


WTAMU > Virtual Math Lab > College Algebra > Tutorial 14: Linear Equations in One Variable


 

 

check markAnswer/Discussion to 1a

problem 1a
 

ad1a

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

*Inverse of mult. by 10 is div. both sides by 10

 
 

If you put -2 back in for x in the original problem you will see that -2 is the solution we are looking for.

 
(return to problem 1a)


 

 

check markAnswer/Discussion to 1b

problem 1b
 

ad1b
*Get all x terms on one side

*Inverse of add. 5 is sub. 5
 

*Inverse of mult. by -1 is div. by -1

 
 

If you put 7 back in for x in the original problem you will see that 7 is the solution we are looking for.

 
(return to problem 1b)


 

 

check markAnswer/Discussion to 1c

problem 1c
 

ad1c
*Remove ( ) by using dist. prop.
*Combine like terms

*Get all x terms on one side

*Inverse of sub. 16 is add. 16
 
 

*Inverse of mult. by 4 is div. by 4
 

 
 

If you put 5/2 back in for x in the original problem you will see that 5/2 is the solution we are looking for.

 
(return to problem 1c)


 

 

check markAnswer/Discussion to 2a

problem 2a
 

ad2a

*Remove ( ) by using dist. prop.
*Combine like terms

*Get all the x terms on one side
 
 

Where did our variable, x, go???  It disappeared on us.  Also note how we ended up with a FALSE statement, -3 is not equal to -4.  This does not mean that x = -3 or x = -4. 

Whenever your variable drops out AND you end up with a FALSE statement, then after all of your hard work, there is NO SOLUTION.

So, the answer is no solution which means this is an inconsistent equation.
 

(return to problem 2a)

 

check markAnswer/Discussion to 2b

problem 2b
 

ad2b

*To get rid of the fractions, 
mult. both sides by the LCD of 4
 
 
 

*Get all the x terms on one side
 

*Inverse of add. 2 is sub. 2
 
 

*Inverse of mult. by -3 is div. by -3
 
 

If you put 4/3 back in for x in the original problem you will see that 4/3  is the solution we are looking for.

This would be an example of a conditional equation, because we came up with one solution.
 

(return to problem 2b)


 

 

Check markAnswer/Discussion to 2c

problem 2c
 

ad2c

*Remove ( ) by using dist. prop.

*Get all the x terms on one side
 
 

Where did our variable, x, go???  It disappeared on us.  Also note how we ended up with a TRUE statement, -27 does indeed equal -27.  This does not mean that x = -27. 

Whenever your variable drops out AND you end up with a TRUE statement, then the solution is ALL REAL NUMBERS. This means that if you plug in any real number for x in this equation, the left side will equal the right side.

So the answer is all real numbers, which means this equation is an identity.
 

(return to problem 2c)

 

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WTAMU > Virtual Math Lab > College Algebra > Tutorial 14: Linear Equations in One Variable


Last revised on Dec. 16, 2009 by Kim Seward.
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