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This is one of our many quizzes on Solving rational equations. Click on this to access more such competitive quizzes on Algebra.

Solving rational equations Quiz 2

mathematics

 Solve for the value of $x : \dfrac{{3x}}{{{x^2} + 5x + 6}} + \dfrac{2}{{{x^2} + x - 2}} = \dfrac{{5x}}{{{x^2} + 2x - 3}}$

 Multiply the whole thing with all the terms of denominator

$  \dfrac{{3x}}{{{x^2} + 5x + 6}} + \dfrac{2}{{{x^2} + x - 2}} = \dfrac{{5x}}{{{x^2} + 2x - 3}} \\  \dfrac{{3x}}{{(x + 2)(x + 3)}} + \dfrac{2}{{(x - 1)(x + 2)}} = \dfrac{{5x}}{{(x - 1)(x + 3)}} \\  (x + 2)(x + 3)(x - 1)(\dfrac{{3x}}{{(x + 2)(x + 3)}} + \dfrac{2}{{(x - 1)(x + 2)}} = \dfrac{{5x}}{{(x - 1)(x + 3)}}) \\  3x(x - 1) + 2(x + 3) = 5x(x + 2) \\  3{x^2} - x + 6 = 5{x^2} + 10x \\  2{x^2} + 11x - 6 = 0 \\  x = \dfrac{1}{2}\&  - 6 \\  $

 Find the value of $x$ in the equation : $\dfrac{{3x - 2}}{{9x + 1}} = \dfrac{{2x - 5}}{{6x - 5}}$

 Just cross multiply the terms with each other.

$  \dfrac{{3x - 2}}{{9x + 1}} = \dfrac{{2x - 5}}{{6x - 5}} \\  (3x - 2)(6x - 5) = (2x - 5)(9x + 1) \\  18{x^2} - 27x + 10 = 18{x^2} - 43x - 5 \\   - 27x + 43x =  - 5 - 10 \\  16x =  - 15 \\  x =  - \dfrac{{15}}{{16}} \\  $

 Find the value of x in the equation:  $\dfrac{{10}}{{x + 4}} = \dfrac{{15}}{{4(x + 1)}}$

 Solve by using the cross multiplication method.

$  \dfrac{{10}}{{x + 4}} = \dfrac{{15}}{{4(x + 1)}} \\  40x + 40 = 15x + 60 \\  25x = 20 \\  x = \dfrac{{20}}{{25}} \\  x = \dfrac{4}{5} \\  $

 Find the number of solution for the equation: $\dfrac{x}{{x - 2}} + \dfrac{1}{{x - 4}} = \dfrac{2}{{{x^2} - 6x + 8}}$

 Make sure for the given answer the value of denominator don’t become 0

$  \dfrac{x}{{x - 2}} + \dfrac{1}{{x - 4}} = \dfrac{2}{{{x^2} - 6x + 8}} \\  \dfrac{x}{{x - 2}} + \dfrac{1}{{x - 4}} = \dfrac{2}{{(x - 2)(x - 4)}} \\  \dfrac{{x(x - 4) + (x - 2)}}{{(x - 2)(x - 4)}} = \dfrac{2}{{(x - 2)(x - 4)}} \\  {x^2} - 4x + x - 2 = 2 \\  {x^2} - 3x - 4 = 0 \\  x = 4 \&  - 1 \\  $ But 4 can’t be the solution as that would make the value of denominator as 0, therefore there are only one answer to the problem

 Find the number of solution for the equation: $\dfrac{{10}}{{x(x - 2)}} + \dfrac{4}{x} = \dfrac{5}{{x - 2}}$

$  \dfrac{{10}}{{x(x - 2)}} + \dfrac{4}{x} = \dfrac{5}{{x - 2}} \\  \dfrac{{10}}{{x(x - 2)}} = \dfrac{5}{{x - 2}} - \dfrac{4}{x} \\  \dfrac{{10}}{{x(x - 2)}} = \dfrac{{5x - 4x + 8}}{{x(x - 2)}} \\  10 - 8 = x \\  x = 2 \\  $ But this can’t be the solution as the value of $x$ can’t be 0 or 2 otherwise the whole equation would be divided by 0 making it infinity. $ \\ $ So the answer would be no solution.