Solve for b given that (2b-3)/(3b+2)=(2b+1)/(3b-2)

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve for b given that: (2b - 3) / (3b + 2)  = (2b + 1) / (3b - 2)

(2b - 3) / (3b + 2)  = (2b + 1) / (3b - 2)

=> (2b - 3)(3b - 2) = (2b + 1)(3b + 2)

=> 6b^2 - 9b - 4b + 6 = 6b^2 + 3b + 4b + 2

=> 6b^2 - 9b - 4b + 6 - 6b^2 - 3b - 4b - 2 = 0

=> -20b + 4 = 0

=> -20b = -4

=> b = -4/(-20)

=> b = 1/5

The required value of b = 1/5

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

Do you mean to solve (2b-3)/(3b+2) = (2b+1)/(3b-2) ?

Cross multiply:

(2b-3)(3b-2) = (2b+1)(3b+2)

6b^2-4b-9b+6 = 6b^2+4b+3b+2

Collect like terms together:

(6-6)b^2+(-4-9-4-3)b+6-2 = 0

-20b+4 = 0

20b = 4

b = 4/20

b = 1/5.

But

2b-3/3b+2=2b+1/3b-2 is different from (2n-3)/(3b+2) = (2b+1)/(3b-2)

We solve 2b-3/3b+2=2b+1/3b-2 as

Collect like terms on one side and nubers on the other side.

2b-2b -3/3b-1/3b = -2-2

-4/3b = -4

3b/4 = 1/4

b =1/3.

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