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Solve equations with substitution method

From Rosetta Code
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.
Task


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


See related



Julia[edit]

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, y
end
 
@show solvetwolinear("3x + y = -1", "2x - 3y = -19") # solvetwolinear("3x + y = -1", "2x - 3y = -19") = (-2.0, 5.0)
 

Perl[edit]

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[edit]

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[1]/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[1] -= 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[edit]

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[edit]

 
firstEquation = [3.0,1.0,-1.0] secondEquation = [2.0,-3.0,-19.0]
getCrossingPoint(firstEquation,secondEquation)
 
func getCrossingPoint(firstEquation,secondEquation)
x1 = firstEquation[1] y1 = firstEquation[2] r1 = firstEquation[3] x2 = secondEquation[1] y2 = secondEquation[2] r2 = secondEquation[3]
temp = []
add(temp,x1) add(temp,-y1) add(temp,r1)
resultY = ((temp[1]* r2) - (x2 * temp[3])) / ((x2 * temp[2]) + (temp[1]*y2)) resultX = (r1 - (y1*resultY)) / x1
see "x = " + resultX + nl + "y = " + resultY + nl
 
Output:
x = -2
y = 5

Wren[edit]

var solve = Fn.new { |e1, e2|
e2 = e2.toList
for (i in 1..2) e2[i] = e2[i] * e1[0] / e2[0]
var y = (e2[2] - e1[2]) / (e2[1] - e1[1])
var x = (e1[2] - e1[1] * y) / e1[0]
return [x, y]
}
 
var e1 = [3, 1, -1]
var e2 = [2, -3, -19]
var sol = solve.call(e1, e2)
System.print("x = %(sol[0]), y = %(sol[1])")
Output:
x = -2, y = 5

XPL0[edit]

This shows the vector routines from xpllib.xpl.

func real VSub(A, B, C);        \Subtract two 3D vectors
real 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 scalar
real 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