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

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

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

We move all terms to the left:

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


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

x^2+x/5-(-1x+7)=0

We get rid of parentheses

x^2+x/5+1x-7=0

We multiply all the terms by the denominator


x^2*5+x+1x*5-7*5=0

We add all the numbers together, and all the variables

x^2*5+x+1x*5-35=0

Wy multiply elements

5x^2+x+5x-35=0

We add all the numbers together, and all the variables

5x^2+6x-35=0

a = 5; b = 6; c = -35;
Δ = b2-4ac
Δ = 62-4·5·(-35)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{46}}{2*5}=\frac{-6-4\sqrt{46}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{46}}{2*5}=\frac{-6+4\sqrt{46}}{10}
$