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

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

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

We move all terms to the left:

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


Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(1/4)x-4+(1/3)=0

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

x^2-4+1/3=0

We multiply all the terms by the denominator


x^2*3+1-4*3=0

We add all the numbers together, and all the variables

x^2*3-11=0

Wy multiply elements

3x^2-11=0

a = 3; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·3·(-11)
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{33}}{2*3}=\frac{0-2\sqrt{33}}{6}
=-\frac{2\sqrt{33}}{6}
=-\frac{\sqrt{33}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{33}}{2*3}=\frac{0+2\sqrt{33}}{6}
=\frac{2\sqrt{33}}{6}
=\frac{\sqrt{33}}{3}
$