Question

With a detailed explanation from the internet 1/(x-5)+3/(x+2)=4

Answer 1

Answer:

$\boxed{ \sf x=\dfrac{4\pm\sqrt{43}}{2}}$

Explanation:

$\rightarrow \sf \dfrac{1}{x-5}+\dfrac{3}{x+2}=4$

make the denominators same

$\rightarrow \sf \dfrac{1(x+2)}{(x-5)(x+2)}+\dfrac{3(x-5)}{(x+2)(x-5)}=4$

join the fractions together

$\rightarrow \sf \dfrac{x+2+3x-15}{(x-5)(x+2)}=4$

cross multiply

$\rightarrow \sf x+2+3x-15=4(x-5)(x+2)$

simplify

$\rightarrow \sf 4x-13=4x^2-12x-40$

group the variables

$\rightarrow \sf 4x^2-16x-27 = 0$

use quadratic formula

$\rightarrow \sf x = \dfrac{-\left(-16\right)\pm \sqrt{\left(-16\right)^2-4\cdot \:4\left(-27\right)}}{2\cdot \:4}$

simplify the following

$\rightarrow \sf x=\dfrac{4\pm\sqrt{43}}{2}$

Answer 2

Answer:

$x=\dfrac{4 \pm \sqrt{43}}{2}$

Step-by-step explanation:

Given equation:

$\dfrac{1}{(x-5)}+\dfrac{3}{(x+2)}=4$

Make the denominators of the algebraic fractions the same, then combine them into one fraction:

$\begin{aligned}\implies \dfrac{1}{(x-5)} \cdot \dfrac{(x+2)}{(x+2)}+\dfrac{3}{(x+2)}\cdot \dfrac{(x-5)}{(x-5)} & =4\\\\\implies \dfrac{x+2}{(x-5)(x+2)}+\dfrac{3(x-5)}{(x-5)(x+2)} & = 4\\\\ \implies \dfrac{x+2+3(x-5)}{(x-5)(x+2)} & = 4\\\\ \implies \dfrac{4x-13}{(x-5)(x+2)} & = 4 \end{aligned}$

Multiply both sides of the equation by $(x-5)(x+2)$:

$\begin{aligned}\implies \dfrac{(4x-13)}{(x-5)(x+2)}\cdot (x-5)(x+2) & = 4(x-5)(x+2)\\\\\dfrac{(4x-13)(x-5)(x+2)}{(x-5)(x+2)} & = 4(x-5)(x+2)\\\\4x-13 & = 4(x-5)(x+2)\\\\4x-13 & =4x^2-12x-40\\\\4x^2-16x-27 & = 0\end{aligned}$

Solve using the Quadratic Formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0$

$\implies a=4\\\implies b=-16\\\implies c=-27$

Therefore:

$\implies x=\dfrac{-(-16) \pm \sqrt{(-16)^2-4(4)(-27)} }{2(4)}$

$\implies x=\dfrac{16 \pm \sqrt{688}}{8}$

$\implies x=\dfrac{16 \pm 4 \sqrt{43}}{8}$

$\implies x=\dfrac{4 \pm \sqrt{43}}{2}$