5/9q+7=4-2/3q

Solution for 5/9q+7=4-2/3q equation:

5/9q+7=4-2/3q

We move all terms to the left:

5/9q+7-(4-2/3q)=0


Domain of the equation: 9q!=0

q!=0/9

q!=0

q∈R



Domain of the equation: 3q)!=0

q!=0/1

q!=0

q∈R


We add all the numbers together, and all the variables

5/9q-(-2/3q+4)+7=0

We get rid of parentheses

5/9q+2/3q-4+7=0

We calculate fractions


15q/27q^2+18q/27q^2-4+7=0

We add all the numbers together, and all the variables

15q/27q^2+18q/27q^2+3=0

We multiply all the terms by the denominator


15q+18q+3*27q^2=0

We add all the numbers together, and all the variables

33q+3*27q^2=0

Wy multiply elements

81q^2+33q=0

a = 81; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·81·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*81}=\frac{-66}{162}
=-11/27
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*81}=\frac{0}{162}
=0
$