R: Automate solving many systems of equations in data frame

I'm working on a dataset that will solve simple equation systems of 2 equations with 2 unknowns. However, as there are hundreds of such systems, I will need to find some algorithm, perhaps a loop, that will perform the task for n rows in a data frame.

Consider the following data frame:

df <- as.data.frame(rbind(c(1,	0.214527122,	166486.7605,	1,	-0.003217907,	0),
            c(1,	0.227828903,	160088.1385,	1,	-0.003417434,	0),
            c(1,	0.214324606,	154445.4132,	1,	-0.003214869,	0),
            c(1,	0.218698883,	147921.278,	    1,	-0.003280483,	0),
            c(1,	0.201444364,	151268.6711,	1,	-0.003021665,	0)))
colnames(df) = c("y1",	"x1",	"b1",	"y2",	"x2",	"b2")

Notice how each row in this data set contains two linear equations, for two unknowns, x and y.

> df
  y1        x1       b1 y2           x2 b2
1  1 0.2145271 166486.8  1 -0.003217907  0
2  1 0.2278289 160088.1  1 -0.003417434  0
3  1 0.2143246 154445.4  1 -0.003214869  0
4  1 0.2186989 147921.3  1 -0.003280483  0
5  1 0.2014444 151268.7  1 -0.003021665  0

The two equations represent supply and demand curves in an economic model. I want to calculate the resulting equilibrium price and quantity for each row in the data frame.

I can, for example, solve the first row with the following code:

A <- rbind(c(df$y1[1], df$x1[1]), 
           c(df$y2[1], df$x2[1]))

B <- c(df$b1[1],df$b2[1])

> solve(A,B)
[1]   2460.396 764595.000

At equilibrium, price is 2460.4 and quantity is 764 595. I can then add this result to the data frame in two new columns:

df$Price[1] <- solve(A,B)[1]
df$Quantity[1] <- solve(A,B)[2]

> df
  y1        x1       b1 y2           x2 b2    Price Quantity
1  1 0.2145271 166486.8  1 -0.003217907  0 2460.396   764595
2  1 0.2278289 160088.1  1 -0.003417434  0       NA       NA
3  1 0.2143246 154445.4  1 -0.003214869  0       NA       NA
4  1 0.2186989 147921.3  1 -0.003280483  0       NA       NA
5  1 0.2014444 151268.7  1 -0.003021665  0       NA       NA

That was the first row. But I haven't figured out a good way of automating this to solve for each row in the data frame. I suppose some type of loop might be an option, but I've had no luck in solving it.

Rearrange the columns so that the first 4 are the matrix and then run the following apply :

df2 <- df[c("y1", "y2", "x1", "x2", "b1", "b2")]
t(apply(df2, 1, function(x) solve(matrix(x[1:4], 2), x[5:6])))

giving

         [,1]   [,2]
[1,] 2460.396 764595
[2,] 2365.835 692284
[3,] 2282.444 709965
[4,] 2186.029 666374
[5,] 2235.497 739823

One dplyr option

library(dplyr)
library(purrr)
df |>
  dplyr::rowwise() |>
  dplyr::mutate(
    Price = (solve(rbind(
      c(cur_data()$y1, cur_data()$x1),
      c(cur_data()$y2, cur_data()$x2)
    ), c(
      cur_data()$b1,
      cur_data()$b2
    )) |>
      pluck(1)
    ),
    Quantity = (solve(rbind(
      c(cur_data()$y1, cur_data()$x1),
      c(cur_data()$y2, cur_data()$x2)
    ), c(
      cur_data()$b1,
      cur_data()$b2
    )) |>
      pluck(2)
    )
  )

Output:

# A tibble: 5 × 8
# Rowwise: 
     y1    x1      b1    y2       x2    b2 Price Quantity
  <dbl> <dbl>   <dbl> <dbl>    <dbl> <dbl> <dbl>    <dbl>
1     1 0.215 166487.     1 -0.00322     0 2460.  764595.
2     1 0.228 160088.     1 -0.00342     0 2366.  692284.
3     1 0.214 154445.     1 -0.00321     0 2282.  709965.
4     1 0.219 147921.     1 -0.00328     0 2186.  666374.
5     1 0.201 151269.     1 -0.00302     0 2235.  739823.

To add two new columns, consider mapply, elementwise processing member of apply family:

get_equilibrium <- function(y1, x1, b1, y2, x2, b2) {
  A <- rbind(c(y1, x1), c(y2, x2))
  B <- c(b1, b2)
  solve(A, B)
}

df[c("Price", "Quantity")] <- t(mapply(
  get_equilibrium, df$y1, df$x1, df$b1, df$y2, df$x2, df$b2
))

df
#   y1        x1       b1 y2           x2 b2    Price Quantity
# 1  1 0.2145271 166486.8  1 -0.003217907  0 2460.396   764595
# 2  1 0.2278289 160088.1  1 -0.003417434  0 2365.835   692284
# 3  1 0.2143246 154445.4  1 -0.003214869  0 2282.444   709965
# 4  1 0.2186989 147921.3  1 -0.003280483  0 2186.029   666374
# 5  1 0.2014444 151268.7  1 -0.003021665  0 2235.497   739823