I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)

# Solve equations with substitution method

Solve equations with substitution method is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Let given equations:
3x + y = -1 and 2x - 3y = -19
Solve it with substitution method.

See related

## Julia

`function parselinear(s)    ab, c = strip.(split(s, "="))    a, by = strip.(split(ab, "x"))    b = replace(by, r"[\sy]" => "")    b[end] in "-+" && (b *= "1")    b = replace(b, "--" => "")    return map(x -> parse(Float64, x == "" ? "1" : x), [a, b, c])end function solvetwolinear(s1, s2)    a1, b1, c1 = parselinear(s1)    a2, b2, c2 = parselinear(s2)    x = (b2 * c1 - b1 * c2) / (b2 * a1 - b1 * a2)    y = (a1 * x - c1 ) / -b1    return x, yend @show solvetwolinear("3x + y = -1", "2x - 3y = -19")  # solvetwolinear("3x + y = -1", "2x - 3y = -19") = (-2.0, 5.0) `

## Perl

`use strict;use warnings;use feature 'say'; sub parse {    my(\$e) = @_;    \$e =~ s/ ([xy])/ 1\$1/;    \$e =~ s/[ =\+]//g;    split /[xy=]/, \$e;} sub solve {    my(\$a1, \$b1, \$c1, \$a2, \$b2, \$c2) = @_;    my \$X = ( \$b2 * \$c1  -  \$b1 * \$c2 )          / ( \$b2 * \$a1  -  \$b1 * \$a2 );    my \$Y = ( \$a1 * \$X  -  \$c1 ) / -\$b1;    return \$X, \$Y;} say my \$result = join ' ', solve( parse('3x + y = -1'), parse('2x - 3y = -19') );`
Output:
`-2 5`

## Phix

Slightly modified copy of solveN() from Solving_coin_problems#Phix, admittedly a tad overkill for this task, as it takes any number of rules and any number of variables.

```with javascript_semantics
procedure solve(sequence rules, unknowns)
--
-- Based on https://mathcs.clarku.edu/~djoyce/ma105/simultaneous.html
--  aka the ancient Chinese Jiuzhang suanshu ~100 B.C. (!!)
--
-- Example:
--  rules = {{18,1,1},{38,1,5}}, ie 18==p+n, 38==p+5*n
--  unknowns = {"pennies","nickels"}
--
--  In the elimination phase, both p have multipliers of 1, so we can
--  ignore those two sq_mul and just do (38=p+5n)-(18=p+n)==>(20=4n).
--  Obviously therefore n is 5 and substituting backwards p is 13.
--
string res
sequence sentences = rules, ri, rj
integer l = length(rules), rii, rji
rules = deep_copy(rules)
for i=1 to l do
-- successively eliminate (grow lower left triangle of 0s)
ri = rules[i]
if length(ri)!=l+1 then ?9/0 end if
rii = ri[i+1]
if rii=0 then ?9/0 end if
for j=i+1 to l do
rj = rules[j]
rji = rj[i+1]
if rji!=0 then
rj = sq_sub(sq_mul(rj,rii),sq_mul(ri,rji))
if rj[i+1]!=0 then ?9/0 end if -- (job done)
rules[j] = rj
end if
end for
end for
for i=l to 1 by -1 do
-- then substitute each backwards
ri = rules[i]
rii = ri/ri[i+1] -- (all else should be 0)
rules[i] = sprintf("%s = %d",{unknowns[i],rii})
for j=i-1 to 1 by -1 do
rj = rules[j]
rji = rj[i+1]
if rji!=0 then
rules[j] = 0
rj -= rji*rii
rj[i+1] = 0
rules[j] = rj
end if
end for
end for
res = join(rules,", ")
printf(1,"%v ==> %s\n",{sentences,res})
end procedure

--for 3x + y = -1 and 2x - 3y = -19:
solve({{-1,3,1},{-19,2,-3}},{"x","y"})
```
Output:
```{{-1,3,1},{-19,2,-3}} ==> x = -2, y = 5
```

Alternatively, since I'm staring right at it, here's a

Translation of: Raku
```with javascript_semantics
function solve2(sequence e1,e2)
atom {a1,b1,c1} = e1,
{a2,b2,c2} = e2,
x = (b2*c1 - b1*c2)
/ (b2*a1 - b1*a2),
y = (a1*x - c1)/-b1;
return {x, y}
end function
printf(1,"x = %d, y = %d\n",solve2({3,1,-1},{2,-3,-19}))
```
Output:
```x = -2, y = 5
```

## Raku

`sub solve-system-of-two-linear-equations ( [ \a1, \b1, \c1 ], [ \a2, \b2, \c2 ] ) {    my \X = ( b2 * c1   -   b1 * c2 )          / ( b2 * a1   -   b1 * a2 );     my \Y = ( a1 * X    -   c1 ) / -b1;     return X, Y;}say solve-system-of-two-linear-equations( (3,1,-1), (2,-3,-19) );`
Output:
`(-2 5)`

## Ring

` firstEquation = [3.0,1.0,-1.0] secondEquation = [2.0,-3.0,-19.0]getCrossingPoint(firstEquation,secondEquation) func getCrossingPoint(firstEquation,secondEquation)     x1 = firstEquation y1 = firstEquation r1 = firstEquation x2 = secondEquation y2 = secondEquation r2 = secondEquation     temp = []     add(temp,x1) add(temp,-y1) add(temp,r1)     resultY = ((temp* r2) - (x2 * temp)) / ((x2 * temp) + (temp*y2)) resultX = (r1 - (y1*resultY)) / x1      see "x = " + resultX + nl + "y = " + resultY + nl `
Output:
```x = -2
y = 5
```

## Wren

`var solve = Fn.new { |e1, e2|    e2 = e2.toList    for (i in 1..2) e2[i] = e2[i] * e1 / e2    var y = (e2 - e1) / (e2 - e1)    var x = (e1 - e1 * y) / e1    return [x, y]} var e1 = [3, 1, -1]var e2 = [2, -3, -19]var sol = solve.call(e1, e2)System.print("x = %(sol), y = %(sol)")`
Output:
```x = -2, y = 5
```

## XPL0

This shows the vector routines from xpllib.xpl.

`func real VSub(A, B, C);        \Subtract two 3D vectorsreal    A, B, C;                \A:= B - C[A(0):= B(0) - C(0);            \VSub(A, A, C) => A:= A - C A(1):= B(1) - C(1); A(2):= B(2) - C(2);return A;];      \VSub func real VMul(A, B, S);        \Multiply 3D vector by a scalarreal    A, B, S;                \A:= B * S[A(0):= B(0) * S;               \VMul(A, A, S) => A:= A * S A(1):= B(1) * S; A(2):= B(2) * S;return A;];      \VMul real E1, E2, X1, X2, X, Y;[E1:= [3.,  1.,  -1.]; E2:= [2., -3., -19.];X1:= E1(0);X2:= E2(0);VMul(E1, E1, X2);VMul(E2, E2, X1);VSub(E1, E1, E2);Y:= E1(2)/E1(1);E2(1):= E2(1)*Y;E2(2):= E2(2)-E2(1);X:= E2(2)/E2(0);Text(0, "x = ");  RlOut(0, X);  CrLf(0);Text(0, "y = ");  RlOut(0, Y);  CrLf(0);]`
Output:
```x =    -2.00000
y =     5.00000
```