If it's not what You are looking for type in the equation solver your own equation and let us solve it.
(1/x+2)+5=16/(x+2)
We move all terms to the left:
(1/x+2)+5-(16/(x+2))=0
Domain of the equation: x+2)!=0
x∈R
Domain of the equation: (x+2))!=0We get rid of parentheses
x∈R
1/x-(16/(x+2))+2+5=0
We calculate fractions
(1*(x+2)))/3x^2+(-(16*x)/3x^2+2+5=0
We add all the numbers together, and all the variables
(1*(x+2)))/3x^2+(-(+16x)/3x^2+2+5=0
We add all the numbers together, and all the variables
(1*(x+2)))/3x^2+(-(+16x)/3x^2+7=0
We calculate fractions
((1*(x+2)))*3x^2)/(3x^2+(*3x^2)+(-(+16x)*3x^2)/(3x^2+(*3x^2)+7=0
We get rid of parentheses
((1*(x+2)))*3x^2)/(3x^2+*3x^2+(-(+16x)*3x^2)/(3x^2+(*3x^2)+7=0
We calculate fractions
*3x^2+(((1*(x+2)))*3x^2)*(3x^2+*3x^2+7)/((3x^2*(3x^2+(*3x^2)+7)+((-(+16x)*3x^2)*3x^2/((3x^2*(3x^2+(*3x^2)+7)=0
We calculate terms in parentheses: +(((1*(x+2)))*3x^2)*(3x^2+*3x^2+7)/((3x^2*(3x^2+(*3x^2)+7)+((-(+16x)*3x^2)*3x^2/((3x^2*(3x^2+(*3x^2)+7), so:
((1*(x+2)))*3x^2)*(3x^2+*3x^2+7)/((3x^2*(3x^2+(*3x^2)+7)+((-(+16x)*3x^2)*3x^2/((3x^2*(3x^2+(*3x^2)+7
We can not solve this equation
See similar equations:
| b/3+2=-1 | | F(7)=a+5 | | 3/x+3=3/2x+13/4 | | 0=50+0,2.t | | F(a)=a+5 | | 8z−5z=18z= | | 3x-5+3=-8 | | (x+30)=3x-40) | | 5-1=2y+8 | | 2/x+3=(3/2x)+25/8 | | |3n+2|=-10 | | 12x+9x+18=180 | | 35-x5=-7 | | t/18=12/8 | | 12x/2+9x+18=180 | | x-12.95=44 | | (x+2/18)=1/9+(x-5/4) | | 5(2x+2)=17 | | -3x+5x=28 | | 7x=52.5 | | 28t-24-7=221 | | -15=3(n+2) | | x/18=12/8 | | 14d-14+5=117 | | 2/5n=-2*123/35 | | (x+2)/18=1/9+(x-5)/4 | | x+1.3=4.3 | | -2=4p | | 117=2(7d−7)+5 | | -6-2k=8 | | 65=65=x+1.5x+2.25x+3.375x | | 2(x+7)=1/2(4x-12) |