If it's not what You are looking for type in the equation solver your own equation and let us solve it.
(5x-3)/(x-1)+3/x=0
Domain of the equation: (x-1)!=0
We move all terms containing x to the left, all other terms to the right
x!=1
x∈R
Domain of the equation: x!=0We calculate fractions
x∈R
(5x^2-3x)/(x^2-1x)+(3x-3)/(x^2-1x)=0
We multiply all the terms by the denominator
(5x^2-3x)+(3x-3)=0
We get rid of parentheses
5x^2-3x+3x-3=0
We add all the numbers together, and all the variables
5x^2-3=0
a = 5; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·5·(-3)
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*5}=\frac{0-2\sqrt{15}}{10} =-\frac{2\sqrt{15}}{10} =-\frac{\sqrt{15}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*5}=\frac{0+2\sqrt{15}}{10} =\frac{2\sqrt{15}}{10} =\frac{\sqrt{15}}{5} $
| 4y+8=-40 | | 3(2m-1)+14=0 | | 26-6(2x-8)=-4(x-2)+4x | | 2x4+5x3-4x2-10x-3=0 | | 9x+3(7-x)=2(x-4) | | (X+2)(x+3)+(x-3)(x-2)-2x(x-1)=0 | | 20+2,5(x+4)=14-(5x+8) | | 2y+2y=36 | | 2y+2y=26 | | 2801.25=2700.00+1350r | | 6x-4=4x-5 | | 3÷4(x-1)=x-3 | | (2x-7)(2x-7)-4=0 | | 4=2a-8 | | 2+3(t+2)=11 | | 25=0.5f+5 | | 7=24+e | | (5-2b)+4-(3b-11)-6=0 | | 5c+10=25 | | (2x-1)^2+1=8+2x | | 3(x-1)=12+4(x-1) | | 9x^2+16/x^2=40 | | 9x^2+16x^-2=40 | | 9x^2+16x^-2-40=0 | | 2x÷5-3÷2=x÷2+1 | | -5x²-8x-56=0 | | X-6/4-x-4/6=x-1/10 | | 4x^2+12x+9=54 | | u^2+6u-24=0 | | 23=5x-8 | | 180=12x^2+40x | | 12x^2+40x=180 |