3/7x+1/3x=48

Solution for 3/7x+1/3x=48 equation:

3/7x+1/3x=48

We move all terms to the left:

3/7x+1/3x-(48)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We calculate fractions


9x/21x^2+7x/21x^2-48=0

We multiply all the terms by the denominator


9x+7x-48*21x^2=0

We add all the numbers together, and all the variables

16x-48*21x^2=0

Wy multiply elements

-1008x^2+16x=0

a = -1008; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-1008)·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*-1008}=\frac{-32}{-2016}
=1/63
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-1008}=\frac{0}{-2016}
=0
$