kx+b=y; px=y^2

Ha introducido [src]

k*x + b = y

$$b + k x = y$$

       2
p*x = y

$$p x = y^{2}$$

p*x = y^2

Respuesta rápida

$$b_{1} = - k x + y$$
=
$$- k x + y$$
=

y - k*x

$$p_{1} = \frac{y^{2}}{x}$$
=
$$\frac{y^{2}}{x}$$
=

y^2/x