Question

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

Answer 1

`\boxed{ \sf x=(4\pm√(43))/(2)}`

Step-by-step explanation:

`\rightarrow \sf (1)/(x-5)+(3)/(x+2)=4`

make the denominators same

`\rightarrow \sf (1(x+2))/((x-5)(x+2))+(3(x-5))/((x+2)(x-5))=4`

join the fractions together

`\rightarrow \sf (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 = (-\left(-16\right)\pm √(\left(-16\right)^2-4\cdot \:4\left(-27\right)))/(2\cdot \:4)`

simplify the following

`\rightarrow \sf x=(4\pm√(43))/(2)`

Answer 2

`x=(4 \pm √(43))/(2)`

Step-by-step explanation:

Given equation:

`(1)/((x-5))+(3)/((x+2))=4`

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

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

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

`\begin{aligned}\implies ((4x-13))/((x-5)(x+2))\cdot (x-5)(x+2) & = 4(x-5)(x+2)\\\\((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=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when}\:ax^2+bx+c=0`

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

Therefore:

`\implies x=(-(-16) \pm √((-16)^2-4(4)(-27)) )/(2(4))`

`\implies x=(16 \pm √(688))/(8)`

`\implies x=(16 \pm 4 √(43))/(8)`

`\implies x=(4 \pm √(43))/(2)`