If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5/2x+7/4=2+5/6x
We move all terms to the left:
5/2x+7/4-(2+5/6x)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: 6x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
5/2x-(5/6x+2)+7/4=0
We get rid of parentheses
5/2x-5/6x-2+7/4=0
We calculate fractions
504x^2/192x^2+480x/192x^2+(-160x)/192x^2-2=0
We multiply all the terms by the denominator
504x^2+480x+(-160x)-2*192x^2=0
Wy multiply elements
504x^2-384x^2+480x+(-160x)=0
We get rid of parentheses
504x^2-384x^2+480x-160x=0
We add all the numbers together, and all the variables
120x^2+320x=0
a = 120; b = 320; c = 0;
Δ = b2-4ac
Δ = 3202-4·120·0
Δ = 102400
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{102400}=320$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-320}{2*120}=\frac{-640}{240} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+320}{2*120}=\frac{0}{240} =0 $
| 3x-(-2)=-9x+22 | | (2/3)z=(5/6)(2+z) | | 17x-8=35 | | -2(3x-2)+3x=34 | | 2x^2+16=-18x | | 8=5q=11+4q | | 15x2+21x–72=0 | | 7+02x=17 | | 5x(3x+2)=280 | | f(25)=1.8(25)+32 | | Y=4x-80 | | 25n=2*25 | | 1/2(3x+5)-1/3(x+7)=6 | | 4(-2x+4)=-2(3x-7) | | 0.15(y-0.2)=20.5(1-y) | | -8b-32+5=12b-3 | | 38=8u+6(u+4) | | 19=-7v+3(v-3) | | 15x2+21x+-72=0 | | 23=4(x+8)+5x | | u^2-12u+27=0 | | 6x−40=260x= | | -4y+6=-2y(4y+15) | | 10u=24+7u | | 8c+6c-c=39 | | -36=4r-8r | | -11+t/2=0 | | -6=-9x+6x | | 8=7(r-2)+1 | | -7(4x+7)=3+4-(4-6x) | | 8(y-7)+10=-86 | | -7b-9b=32 |