If it's not what You are looking for type in the equation solver your own equation and let us solve it.
3/x-10+1/3=9/3x-30
We move all terms to the left:
3/x-10+1/3-(9/3x-30)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: 3x-30)!=0We get rid of parentheses
x∈R
3/x-9/3x+30-10+1/3=0
We calculate fractions
81x/27x^2+(-9x)/27x^2+x/27x^2+30-10=0
We add all the numbers together, and all the variables
81x/27x^2+(-9x)/27x^2+x/27x^2+20=0
We multiply all the terms by the denominator
81x+(-9x)+x+20*27x^2=0
We add all the numbers together, and all the variables
82x+(-9x)+20*27x^2=0
Wy multiply elements
540x^2+82x+(-9x)=0
We get rid of parentheses
540x^2+82x-9x=0
We add all the numbers together, and all the variables
540x^2+73x=0
a = 540; b = 73; c = 0;
Δ = b2-4ac
Δ = 732-4·540·0
Δ = 5329
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{5329}=73$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(73)-73}{2*540}=\frac{-146}{1080} =-73/540 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(73)+73}{2*540}=\frac{0}{1080} =0 $
| 1x^2+12x=35 | | -14+9z=0 | | 10z-5+3z=88 | | x^2-(8/5)x+(7/10)=0 | | 8x+1=~31 | | j+16=35 | | (6x-5)°=180 | | 5-(-1.6x)=-7.8 | | 52=-2(6y+4) | | 4(x/4+3)=-6 | | 3(4x+1)=3x | | 2(x-4)=2(x=3) | | x+215=x+15 | | 5(1-2(2z-1))=-3(3z-1)+1 | | 2(4-3x)=-(x+5)= | | -18=6+8(n+7) | | 63+25x=400+25x | | 8(4v-1)-12v=111(2v-6) | | 1/2(6-4x)=x+21 | | 6/7+12t=3/14-15t | | 8x-3-5=36 | | 0.45=p/14 | | -11•0.5a=22 | | -5x-4(3-4x)=4(x-4)-17 | | 7(1-2n)=-14 | | -11(0.5•a)=22 | | 5-2r+1=3r-5(r-3) | | (x-7)^2=4 | | 3x+6-2x=x+6 | | -8=-(1/4)m | | 8y×2=144 | | -2b-2=3b+8 |