1/3x+2+4/7x+8=48

Solution for 1/3x+2+4/7x+8=48 equation:

1/3x+2+4/7x+8=48

We move all terms to the left:

1/3x+2+4/7x+8-(48)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R


We add all the numbers together, and all the variables

1/3x+4/7x-38=0

We calculate fractions


7x/21x^2+12x/21x^2-38=0

We multiply all the terms by the denominator


7x+12x-38*21x^2=0

We add all the numbers together, and all the variables

19x-38*21x^2=0

Wy multiply elements

-798x^2+19x=0

a = -798; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·(-798)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*-798}=\frac{-38}{-1596}
=1/42
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*-798}=\frac{0}{-1596}
=0
$