(3/7)x+(3/8)=3

Solution for (3/7)x+(3/8)=3 equation:

(3/7)x+(3/8)=3

We move all terms to the left:

(3/7)x+(3/8)-(3)=0


Domain of the equation: 7)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(3/7)x-3+(3/8)=0

We add all the numbers together, and all the variables

(+3/7)x-3+(+3/8)=0

We multiply parentheses

3x^2-3+(+3/8)=0

We get rid of parentheses

3x^2-3+3/8=0

We multiply all the terms by the denominator


3x^2*8+3-3*8=0

We add all the numbers together, and all the variables

3x^2*8-21=0

Wy multiply elements

24x^2-21=0

a = 24; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·24·(-21)
Δ = 2016
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2016}=\sqrt{144*14}=\sqrt{144}*\sqrt{14}=12\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{14}}{2*24}=\frac{0-12\sqrt{14}}{48}
=-\frac{12\sqrt{14}}{48}
=-\frac{\sqrt{14}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{14}}{2*24}=\frac{0+12\sqrt{14}}{48}
=\frac{12\sqrt{14}}{48}
=\frac{\sqrt{14}}{4}
$