5/8x-1=1+7/10x

Solution for 5/8x-1=1+7/10x equation:

5/8x-1=1+7/10x

We move all terms to the left:

5/8x-1-(1+7/10x)=0


Domain of the equation: 8x!=0

x!=0/8

x!=0

x∈R



Domain of the equation: 10x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

5/8x-(7/10x+1)-1=0

We get rid of parentheses

5/8x-7/10x-1-1=0

We calculate fractions


50x/80x^2+(-56x)/80x^2-1-1=0

We add all the numbers together, and all the variables

50x/80x^2+(-56x)/80x^2-2=0

We multiply all the terms by the denominator


50x+(-56x)-2*80x^2=0

Wy multiply elements

-160x^2+50x+(-56x)=0

We get rid of parentheses

-160x^2+50x-56x=0

We add all the numbers together, and all the variables

-160x^2-6x=0

a = -160; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-160)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-160}=\frac{0}{-320}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-160}=\frac{12}{-320}
=-3/80
$