What are possible a values if  l 2a-5 l - l 3a +3 l = 0

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

l 2a-5 l - l 3a +3 l = 0

First we will move l 3a+3l to the right side.

==> l 2a-5 l = l 3a + 3 l

Now we have 4 cases.

Case(1):

(2a-5) = ( 3a+ 3)

We will combine like terms.

==> -a = 8

==> a= 8

 

Case(2):

-(2a-5) = ( 3a+3)

==> -2a + 5 = 3a + 3

==> -2a - 3a = 3 - 5

==> -5a = -2

==> a= 2/5

 

Case(3):

-(2a-5) = - (3a+3)

==> -2a + 5 = -3a - 3

==> -2a + 3a = -3 -5

==> a = -8

 

Case(4):

(2a-5) = - ( 3a+3)

==> 2a - 5 = -3a - 3

==> 2a + 3a = -3 + 5

==> 5a = 2

==> a= 2/5

Then we conclude that all possible a values are :

a = { 2/5 , -8}

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the absolute values for each term:

l 2a-5 l = 2a - 5 for 2a - 5>=0

2a >= 5

a >= 5/2

l 2a-5 l = 5 - 2a for 2a - 5 < 0

a < 5/2

l 3a +3 l =  3a +3 for 3a +3 > = 0

3a >= -3

a >= -1

l 3a +3 l =  -3a -3

a < -1

We'll solve the equation considering 3 ranges of values for a:

1) a is in the range (- infinite -1)

5 - 2a - (-3a -3) = 0

We'll remove the brackets:

5 - 2a + 3a + 3 = 0

We'll combine like terms:

a + 8 = 0

a = -8

Since -8 is in the range of admissible values, we'll accept it.

2) a is in the range (- 1 , 5/2)

5 - 2a - 3a - 3 = 0

We'll combine like terms:

-5a = -2

a = 2/5

Since 2/5 is in the range of admissible values, we'll accept it.

3) a is in the range [5/2 , +infinite)

2a - 5 - 3a - 3 = 0

We'll combine like terms:

a = 8

Since 8 is in the range of admissible values, we'll accept it.

The admissible values for a are: {-8 ; 2/5 ; 8}.

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