1/4x=1/7x+33

Solution for 1/4x=1/7x+33 equation:

1/4x=1/7x+33

We move all terms to the left:

1/4x-(1/7x+33)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 7x+33)!=0

x∈R


We get rid of parentheses

1/4x-1/7x-33=0

We calculate fractions


7x/28x^2+(-4x)/28x^2-33=0

We multiply all the terms by the denominator


7x+(-4x)-33*28x^2=0

Wy multiply elements

-924x^2+7x+(-4x)=0

We get rid of parentheses

-924x^2+7x-4x=0

We add all the numbers together, and all the variables

-924x^2+3x=0

a = -924; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-924)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-924}=\frac{-6}{-1848}
=1/308
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-924}=\frac{0}{-1848}
=0
$