Answer :
I don't know, I will solve for y
1/y+9-2 is 1 so that prpbably means parenthasees
(1/y+9)-2=1/2 times (y+9)
multiply both sides by 2
2(1/y+9)-4=y+9
add 4 to both sides
2(1/y+9)=y+13
distribute
2/y+18=y+13
subtract 13 from both sides
2/y+5=y
multily both sides by y
2+5y=y^2
subtract 2+5y from both sides
y^2-5y-2=0
if we can factor then assume
if you can factor equation into zy=0 then assume z or/and y=0 so
use quadratic formula
in ay^2+by+c=0 then
y=[tex] \frac{-b+ \sqrt{b^2-4ac} }{2a} [/tex] or [tex] \frac{-b- \sqrt{b^2-4ac} }{2a} [/tex]
so subsitute and get
y= [tex] \frac{5+ \sqrt{33} }{2} [/tex] or [tex] \frac{5- \sqrt{33} }{2} [/tex]
1/y+9-2 is 1 so that prpbably means parenthasees
(1/y+9)-2=1/2 times (y+9)
multiply both sides by 2
2(1/y+9)-4=y+9
add 4 to both sides
2(1/y+9)=y+13
distribute
2/y+18=y+13
subtract 13 from both sides
2/y+5=y
multily both sides by y
2+5y=y^2
subtract 2+5y from both sides
y^2-5y-2=0
if we can factor then assume
if you can factor equation into zy=0 then assume z or/and y=0 so
use quadratic formula
in ay^2+by+c=0 then
y=[tex] \frac{-b+ \sqrt{b^2-4ac} }{2a} [/tex] or [tex] \frac{-b- \sqrt{b^2-4ac} }{2a} [/tex]
so subsitute and get
y= [tex] \frac{5+ \sqrt{33} }{2} [/tex] or [tex] \frac{5- \sqrt{33} }{2} [/tex]
2(¹/y+9)-4=y+9
2/y+18=y+9+4
2/y+18=y+13
2/y+[18-13]=y
2/y+5=y
2+5y=y²
y²-5y-2
Y=-b±√b²-4ac
------------------
2a
y=5±√25-4•1•-2
---------------
2
Y=[5+√33]/2
Y=[5-√33]/2
2/y+18=y+9+4
2/y+18=y+13
2/y+[18-13]=y
2/y+5=y
2+5y=y²
y²-5y-2
Y=-b±√b²-4ac
------------------
2a
y=5±√25-4•1•-2
---------------
2
Y=[5+√33]/2
Y=[5-√33]/2
