1/3x+7=3/4x-4

Solution for 1/3x+7=3/4x-4 equation:

1/3x+7=3/4x-4

We move all terms to the left:

1/3x+7-(3/4x-4)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 4x-4)!=0

x∈R


We get rid of parentheses

1/3x-3/4x+4+7=0

We calculate fractions


4x/12x^2+(-9x)/12x^2+4+7=0

We add all the numbers together, and all the variables

4x/12x^2+(-9x)/12x^2+11=0

We multiply all the terms by the denominator


4x+(-9x)+11*12x^2=0

Wy multiply elements

132x^2+4x+(-9x)=0

We get rid of parentheses

132x^2+4x-9x=0

We add all the numbers together, and all the variables

132x^2-5x=0

a = 132; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·132·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*132}=\frac{0}{264}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*132}=\frac{10}{264}
=5/132
$