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

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

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

We move all terms to the left:

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


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 7x+3)!=0

x∈R


We get rid of parentheses

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

We calculate fractions


21x/28x^2+(-20x)/28x^2-3-2=0

We add all the numbers together, and all the variables

21x/28x^2+(-20x)/28x^2-5=0

We multiply all the terms by the denominator


21x+(-20x)-5*28x^2=0

Wy multiply elements

-140x^2+21x+(-20x)=0

We get rid of parentheses

-140x^2+21x-20x=0

We add all the numbers together, and all the variables

-140x^2+x=0

a = -140; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-140)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-140}=\frac{-2}{-280}
=1/140
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-140}=\frac{0}{-280}
=0
$