When a regression is calculated with a simple linear model that returns intercept and slope for an equation like this $y=a + bx$ one can predict $y$, the response variable, based on that equation. Equally one could rearrange for $x$: $x=\frac{\left(y-a\right)}{b}$ and calculate the value of $x$. This isn't available in R's predict() function but can easily be done. Can one do this calculation and still be statistically sound?

When a regression is calculated with a simple linear model that returns intercept and slope for an equation like this $y=a + bx$ one can predict $y$, the response variable, based on that equation. 

Equally one could rearrange for $x$: $x=\frac{\left(y-a\right)}{b}$ and calculate the value of $x$. This isn't available in R's predict() function but can easily be done. Can one do this calculation and still be statistically sound?

When a regression is calculated with a simple linear model that returns intercept and slope for an equation like this y=a + bx one can predict y (the response variable) based on that equation. Equally one could rearrange for x; x={y-a}/b and calculate the value of x. This isn't available in R's predict() function but can easily be done. Can one do this calculation and still be statistically sound?

When a regression is calculated with a simple linear model that returns intercept and slope for an equation like this $y=a + bx$ one can predict $y$, the response variable, based on that equation. Equally one could rearrange for $x$: $x=\frac{\left(y-a\right)}{b}$ and calculate the value of $x$. This isn't available in R's predict() function but can easily be done. Can one do this calculation and still be statistically sound?