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# How do you solve $\dfrac{{7x}}{{2x + 5}} + 1 = \dfrac{{10x - 3}}{{3x}}$?

Last updated date: 18th Jun 2024
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Hint: Here we need to solve for ‘x’. After taking LCM and cross multiplying we will have a quadratic equation. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation.

Complete step-by-step solution:
Given,
$\dfrac{{7x}}{{2x + 5}} + 1 = \dfrac{{10x - 3}}{{3x}}$
Taking LCM on the left hand side of the equation we have,
$\dfrac{{7x + 2x + 5}}{{2x + 5}} = \dfrac{{10x - 3}}{{3x}}$
$\dfrac{{9x + 5}}{{2x + 5}} = \dfrac{{10x - 3}}{{3x}}$
Cross multiplying we have,
$3x(9x + 5) = (2x + 5)(10x - 3)$.
$27{x^2} + 15x = 2x(10x - 3) + 5(10x - 3)$
$27{x^2} + 15x = 20{x^2} - 6x + 50x - 15$
$27{x^2} + 15x - 20{x^2} + 6x - 50x + 15 = 0$
$27{x^2} - 20{x^2} + 15x + 6x - 50x + 15 = 0$
$7{x^2} - 29x + 15 = 0$
On comparing the given equation with the standard quadratic equation $a{x^2} + bx + c = 0$, we have $a = 7$,$b = - 29$ and $c = 15$.
We cannot use the factorization method here, we are unable to split the middle term.
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
$\Rightarrow x = \dfrac{{ - ( - 29) \pm \sqrt {{{( - 29)}^2} - 4(7)(15)} }}{{2(7)}}$
$= \dfrac{{29 \pm \sqrt {841 - 420} }}{{14}}$
$= \dfrac{{29 \pm \sqrt {421} }}{{14}}$
$\Rightarrow x = \dfrac{{29 + \sqrt {421} }}{{14}}$ and $x = = \dfrac{{29 - \sqrt {421} }}{{14}}$.
These are the solutions of $\dfrac{{7x}}{{2x + 5}} + 1 = \dfrac{{10x - 3}}{{3x}}$.