2/3x+3/4=-7+3/5x

Solution for 2/3x+3/4=-7+3/5x equation:

2/3x+3/4=-7+3/5x

We move all terms to the left:

2/3x+3/4-(-7+3/5x)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2/3x-(3/5x-7)+3/4=0

We get rid of parentheses

2/3x-3/5x+7+3/4=0

We calculate fractions


225x^2/240x^2+160x/240x^2+(-144x)/240x^2+7=0

We multiply all the terms by the denominator


225x^2+160x+(-144x)+7*240x^2=0

Wy multiply elements

225x^2+1680x^2+160x+(-144x)=0

We get rid of parentheses

225x^2+1680x^2+160x-144x=0

We add all the numbers together, and all the variables

1905x^2+16x=0

a = 1905; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·1905·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*1905}=\frac{-32}{3810}
=-16/1905
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*1905}=\frac{0}{3810}
=0
$