# using Reduce to solve a system of an equation and an inequality

   Manipulate[Module[{a, b, β, v},
Reduce[{a/b^2 + 1/a == (2 b)/a^2 + 2/3 β b && (3 v)/(
4 π a^2) <b < (3 v)/(2 π a^2)}, b, Reals] // N],
{β, 50, 200}, {v, 1*10^-9, 10*10^-9}
]


I was trying to solve the equation in the Reduce function for a, so I get a(b).Then put the result in the inequality in Reduce to get the numerical range of b. Also I wanted to manipulate beta and v to see how the range of b changes accordingly. But my outcome shows some crazy answer with like 10 || (or).

{β ((a$9202 < 0. && ((β$9202 < (
0.0625 (-48. + 12. a$9202 + 3. a$9202^3))/a$9202^2 && v$9202 > 0. &&
b$9202 == Root[-3 a$9202^3 -
3 a$9202 #1^2 + (6 + 2 a$9202^2 β$9202) #1^3 &, 1]) || ((0.0625 (-48. + 12. a$9202 + 3. a$9202^3))/ a$9202^2 < β$9202 < -(3./a$9202^2) && v$9202 < 0. && b$9202 ==
Root[-3 a$9202^3 - 3 a$9202 #1^2 + (6 + 2 a$9202^2 β$9202) #1^3 &, 1])||


(A part of the outcome I got)

Why doen't it give me one sth<b<sth cleanly? beta and v are constants that I can choose with manipulate and the first equality gives me the value of a with respect to b. (btw what do those \$s and #s mean?)

• Don't use Module[] here. – Feyre Jan 8 '17 at 11:30
• ((3 v) b)/(4 \[Pi] a^2) < (3 v)/(2 \[Pi] a^2) reduces to b < 2 && a != 0 are you sure this is correct? – Feyre Jan 8 '17 at 11:43
• Thank you for replying. ohh oops. I missed one < before b. I edited my question. – Jun Jan 8 '17 at 12:29