In this quadratic equation problem, there are two quantities in algebraic fraction form and a quantity in numerical fraction form. The two fractions in algebraic form are added on the left hand side of the equation and the sum of them is equal to the fraction in the numerical form as displayed on right hand side of the equation.

$\dfrac{2x}{x-4}$ $+$ $\dfrac{2x-5}{x-3}$ $\,=\,$ $\dfrac{25}{3}$

Let’s learn how to solve this algebraic equation to find the value of $x$ mathematically.

There are two fractions on the left-hand side of the equation and they are connected by a plus sign. So, find the addition of the fractions.

$\implies$ $\dfrac{2x \times (x-3)+(x-4) \times (2x-5)}{(x-4)(x-3)}$ $\,=\,$ $\dfrac{25}{3}$

Now, let us focus on simplifying the algebraic expression in the numerator. The product of the factors in each term of the algebraic expression can be calculated by the distributive property.

$\implies$ $\dfrac{2x \times x+2x \times (-3)+x \times (2x-5)-4 \times (2x-5)}{(x-4)(x-3)}$ $\,=\,$ $\dfrac{25}{3}$

$\implies$ $\dfrac{2x^2-6x+x \times 2x+x \times (-5)-4 \times 2x-4 \times (-5)}{(x-4)(x-3)}$ $\,=\,$ $\dfrac{25}{3}$

$\implies$ $\dfrac{2x^2-6x+2x^2-5x-8x+20}{(x-4)(x-3)}$ $\,=\,$ $\dfrac{25}{3}$

$\implies$ $\dfrac{2x^2+2x^2-6x-5x-8x+20}{(x-4)(x-3)}$ $\,=\,$ $\dfrac{25}{3}$

It is time to add the like terms to find the sum of them in the numerator.

$\implies$ $\dfrac{4x^2-19x+20}{(x-4)(x-3)}$ $\,=\,$ $\dfrac{25}{3}$

Similarly, multiply the binomials in the denominator by using the multiplication of algebraic functions.

$\implies$ $\dfrac{4x^2-19x+20}{x \times (x-3)-4 \times (x-3)}$ $\,=\,$ $\dfrac{25}{3}$

$\implies$ $\dfrac{4x^2-19x+20}{x \times x+x \times (-3)-4 \times x-4 \times (-3)}$ $\,=\,$ $\dfrac{25}{3}$

$\implies$ $\dfrac{4x^2-19x+20}{x^2-3x-4x+12}$ $\,=\,$ $\dfrac{25}{3}$

$\implies$ $\dfrac{4x^2-19x+20}{x^2-7x+12}$ $\,=\,$ $\dfrac{25}{3}$

It is time simplify the algebraic equation for preparing the algebraic equation in simplified form.

$\implies$ $3 \times (4x^2-19x+20)$ $\,=\,$ $25 \times (x^2-7x+12)$

$\implies$ $12x^2$ $-$ $57x$ $+$ $60$ $\,=\,$ $25x^2-175x+300$

$\implies$ $25x^2-175x+300$ $\,=\,$ $12x^2$ $-$ $57x$ $+$ $60$

$\implies$ $25x^2$ $-$ $175x$ $+$ $300$ $-$ $12x^2$ $+$ $57x$ $-$ $60$ $\,=\,$ $0$

$\implies$ $25x^2$ $-$ $12x^2$ $-$ $175x$ $+$ $57x$ $+$ $300$ $-$ $60$ $\,=\,$ $0$

$\implies$ $13x^2$ $-$ $118x$ $+$ $240$ $\,=\,$ $0$

The given algebraic equation is successfully simplified as a quadratic equation.

The quadratic equation $13x^2-118x+240$ $\,=\,$ $0$ can be solved by using the quadratic formula.

$x$ $\,=\,$ $\dfrac{-(-118) \pm \sqrt{(-118)^2-4 \times 13 \times 240}}{2 \times 13}$

$\implies$ $x$ $\,=\,$ $\dfrac{118 \pm \sqrt{13924-12480}}{26}$

$\implies$ $x$ $\,=\,$ $\dfrac{118 \pm \sqrt{1444}}{26}$

$\implies$ $x$ $\,=\,$ $\dfrac{118 \pm 38}{26}$

$\implies$ $x$ $\,=\,$ $\dfrac{118+38}{26}$ or $x$ $\,=\,$ $\dfrac{118-38}{26}$

$\implies$ $x$ $\,=\,$ $\dfrac{156}{26}$ or $x$ $\,=\,$ $\dfrac{80}{26}$

$\implies$ $x$ $\,=\,$ $\dfrac{\cancel{156}}{\cancel{26}}$ or $x$ $\,=\,$ $\dfrac{\cancel{80}}{\cancel{26}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $6$ or $x$ $\,=\,$ $\dfrac{40}{13}$

Therefore, the roots or zeros of the given algebraic equation are $x \,=\, 6$ or $x \,=\, \dfrac{40}{13}$.

The solution set for the given algebraic equation is $\bigg\{6, \dfrac{40}{13}\bigg\}$

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