Explanation:
We need to rearrange the following formula for the values given in parenthesis.
(1) x+xy = y, (x)
taking x common in LHS,
x(1+y)=y
[tex]x=\dfrac{y}{1+y}[/tex]
(2) x+y = xy, (x)
Subrtacting both sides by xy.
x+y-xy = xy-xy
x+y-xy = 0
x-xy=-y
x(1-y)=-y
[tex]x=\dfrac{-y}{1-y}[/tex]
(3) x = y+xy, (x)
Subrating both sides by xy
x-xy = y+xy-xy
x(1-y)=y
[tex]x=\dfrac{y}{1-y}[/tex]
(4) E = (1/2)mv^2-(1/2)mu^2, (u)
Subtracting both sides by (1/2)mv^2
E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2
E-(1/2)mv^2 =-(1/2)mu^2
So,
[tex]2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}[/tex]
(5) (x^2/a^2)-(y^2/b^2) = 1, (y)
[tex]\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}[/tex]
(6) ay^2 = x^3, (y)
[tex]y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}[/tex]
Hence, this is the required solution.
