1/5x+13+x=1-9x+22=10

Solution for 1/5x+13+x=1-9x+22=10 equation:

1/5x+13+x=1-9x+22=10

We move all terms to the left:

1/5x+13+x-(1-9x+22)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We add all the numbers together, and all the variables

1/5x+x-(-9x+23)+13=0

We add all the numbers together, and all the variables

x+1/5x-(-9x+23)+13=0

We get rid of parentheses

x+1/5x+9x-23+13=0

We multiply all the terms by the denominator


x*5x+9x*5x-23*5x+13*5x+1=0

Wy multiply elements

5x^2+45x^2-115x+65x+1=0

We add all the numbers together, and all the variables

50x^2-50x+1=0

a = 50; b = -50; c = +1;
Δ = b2-4ac
Δ = -502-4·50·1
Δ = 2300
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:

$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-10\sqrt{23}}{2*50}=\frac{50-10\sqrt{23}}{100}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+10\sqrt{23}}{2*50}=\frac{50+10\sqrt{23}}{100}
$