(1/2x)+(1/5x)=7

Solution for (1/2x)+(1/5x)=7 equation:

(1/2x)+(1/5x)=7

We move all terms to the left:

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


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 5x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


5x/10x^2+2x/10x^2-7=0

We multiply all the terms by the denominator


5x+2x-7*10x^2=0

We add all the numbers together, and all the variables

7x-7*10x^2=0

Wy multiply elements

-70x^2+7x=0

a = -70; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-70)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-70}=\frac{-14}{-140}
=1/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-70}=\frac{0}{-140}
=0
$