Solve for x (2x+3)/(x+2)=(x-1)/(x-2)

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve for x, given (2x+3)/(x+2)=(x-1)/(x-2)

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

=> (2x+3)* ( x - 2) =(x-1)*(x +2)

=> 2x^2 + 3x - 4x - 6 = x^2 + 2x - x - 2

=> 2x^2 + 3x - 4x - 6 - x^2 - 2x + x + 2 = 0

=> x^2 -2x - 4 =0

x1 = [-b + sqrt ( b^2 - 4ac)]/2a

=> [ 2 + sqrt( 4 + 16)]/2

=> [2 + sqrt 20]/ 2

=> 1 + sqrt 5

x1 = [-b - sqrt ( b^2 - 4ac)]/2a

=> [ 2 - sqrt( 4 + 16)]/2

=> [2 - sqrt 20]/ 2

=> 1 - sqrt 5

Therefore x = 1- sqrt 5 and 1+ sqrt 5

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll impose constraints of existence of the given ratios: their denominators must be different from zero.

The range of admissible values for x is R - {-2 ; +2}.

We'll solve the equation by cross multiplying:

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

We'll remove the brackets:

2x^2 - 4x + 3x - 6 = x^2 - x + 2x - 2

We'll move all terms to the left side:

x^2 - 2x - 6 + 2 = 0

x^2 - 2x - 4 = 0

We'll apply quadratic formula:

x1 = [2+sqrt(4 + 16)]/2

x1 = [2+2sqrt5]/2

x1 = 1 + sqrt5

x2 = 1 - sqrt5

Since the solutions are real and different from the excepted values, we'll validate them.

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