(13+2x)/(4x+1)=3/4X

Solution for (13+2x)/(4x+1)=3/4X equation:

(13+2x)/(4x+1)=3/4x

We move all terms to the left:

(13+2x)/(4x+1)-(3/4x)=0


Domain of the equation: (4x+1)!=0

We move all terms containing x to the left, all other terms to the right

4x!=-1

x!=-1/4

x!=-1/4

x∈R



Domain of the equation: 4x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(2x+13)/(4x+1)-(+3/4x)=0

We get rid of parentheses

(2x+13)/(4x+1)-3/4x=0

We calculate fractions


(8x^2+52x)/(16x^2+4x)+(-12x-3)/(16x^2+4x)=0

We multiply all the terms by the denominator


(8x^2+52x)+(-12x-3)=0

We get rid of parentheses

8x^2+52x-12x-3=0

We add all the numbers together, and all the variables

8x^2+40x-3=0

a = 8; b = 40; c = -3;
Δ = b2-4ac
Δ = 402-4·8·(-3)
Δ = 1696
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1696}=\sqrt{16*106}=\sqrt{16}*\sqrt{106}=4\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{106}}{2*8}=\frac{-40-4\sqrt{106}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{106}}{2*8}=\frac{-40+4\sqrt{106}}{16}
$