(4/5)x+(1/7)=3

Solution for (4/5)x+(1/7)=3 equation:

(4/5)x+(1/7)=3

We move all terms to the left:

(4/5)x+(1/7)-(3)=0


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


determiningTheFunctionDomain
(4/5)x-3+(1/7)=0

We add all the numbers together, and all the variables

(+4/5)x-3+(+1/7)=0

We multiply parentheses

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

We get rid of parentheses

4x^2-3+1/7=0

We multiply all the terms by the denominator


4x^2*7+1-3*7=0

We add all the numbers together, and all the variables

4x^2*7-20=0

Wy multiply elements

28x^2-20=0

a = 28; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·28·(-20)
Δ = 2240
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:

$\sqrt{\Delta}=\sqrt{2240}=\sqrt{64*35}=\sqrt{64}*\sqrt{35}=8\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{35}}{2*28}=\frac{0-8\sqrt{35}}{56}
=-\frac{8\sqrt{35}}{56}
=-\frac{\sqrt{35}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{35}}{2*28}=\frac{0+8\sqrt{35}}{56}
=\frac{8\sqrt{35}}{56}
=\frac{\sqrt{35}}{7}
$