7x+3=(1/3)x-4

Solution for 7x+3=(1/3)x-4 equation:

7x+3=(1/3)x-4

We move all terms to the left:

7x+3-((1/3)x-4)=0


Domain of the equation: 3)x-4)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

7x-((+1/3)x-4)+3=0

We multiply all the terms by the denominator


7x*3)x-4)-((+3*3)x-4)+1=0

We add all the numbers together, and all the variables

7x*3)x-4)-(9x-4)+1=0

Wy multiply elements

21x^2=0

a = 21; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·21·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{42}=0$