One Variable Equation

$\begin{aligned} \frac 4{2x+3} + \frac {15}{5x+4} & = \frac {35}{7x+6} & \small \color{#3D99F6} \text{Multiplying both sides by } 7x+6 \\ \frac {28x+24}{2x+3} + \frac {105x+90}{5x+4} & = 35 \\ \frac {14(2x+3)-18}{2x+3} + \frac {21(5x+84)+6}{5x+4} & = 35 \\ 14 - \frac {18}{2x+3} + 21 + \frac 6{5x+4} & = 35 \\ \frac 6{5x+4} & = \frac {18}{2x+3} \\ \frac 1{5x+4} & = \frac 3{2x+3} \\ 15x+12 & = 2x + 3 \\ 13x & = -9 \\ \implies x & = - \frac 9{13} \approx \boxed{-0.692} \end{aligned}$

$\dfrac{4}{2x+3} + \dfrac{15}{5x+4} = \dfrac{35}{7x+6}$

$\dfrac{20x+16+30x+45}{10x^2 + 8x+15x+12} = \dfrac{35}{7x+6}$

$\dfrac{50x+61}{10x^2+23x+12} = \dfrac{35}{7x+6}$

$350x^2 + 300x+427x+366 = 350x^2 + 805x + 420$

$78x + 54 = 0 \Rightarrow \color{#3D99F6}{ x= - 0.692}$

Multiply both sides of the equation by the $LCD$ , but of it is hard to find the $LCD$ , multiply both sides of the equation by the product of the denominators. So, we have

$[\dfrac{4}{2x+3} + \dfrac{15}{5x+4} = \dfrac{35}{7x+6}]$ $(2x+3)(5x+4)(7x+6)$

$4(5x+4)(7x+6)+15(2x+3)(7x+6)=35(2x+3)(5x+4)$

$4(35x^2+30x+28x+24)+15(14x^2+12x+21x+18)=35(10x^2+23x+12)$

$4(35x^2+58x+24)+15(14x^2+33x+18)=35(10x^2+23x+12)$

$140x^2+232x+96+210x^2+495x+270=350x^2+805x+420$

$350x^2+727x+366=350x^2+805x+420$

$366-420=805x-727x$

$-54=78x$

$\boxed{\dfrac{-9}{13} = x}$ $\color{#D61F06}answer$