Solve equations with substitution method

From Rosetta Code
Revision as of 15:32, 28 August 2022 by Thundergnat (talk | contribs) (syntax highlighting fixup automation)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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


AWK[edit]

# syntax: GAWK -f SOLVE_EQUATIONS_WITH_SUBSTITUTION_METHOD.AWK
BEGIN {
    main("3,1,-1","2,-3,-19")
    exit(0)
}
function main(s1,s2,  arr,e1,e2,result_x,result_y,r1,r2,x1,x2,y1,y2) {
    split(s1,e1,",")
    split(s2,e2,",")
    x1 = e1[1]
    y1 = e1[2]
    r1 = e1[3]
    x2 = e2[1]
    y2 = e2[2]
    r2 = e2[3]
    arr[1] = x1
    arr[2] = -y1
    arr[3] = r1
    result_y = ((arr[1]*r2) - (x2*arr[3])) / ((x2*arr[2]) + (arr[1]*y2))
    result_x = (r1 - (y1*result_y)) / x1
    printf("x = %g\ny = %g\n",result_x,result_y)
}
Output:
x = -2
y = 5

BASIC[edit]

BASIC256[edit]

Translation of: FreeBASIC
arraybase 1
dim firstEquation(3)
firstEquation[1] = 3
firstEquation[2] = 1
firstEquation[3] = -1
dim secondEquation(3)
secondEquation[1] = 2
secondEquation[2] = -3
secondEquation[3] = -19

subroutine getCrossingPoint(firstEquation, secondEquation)
	x1 = firstEquation[1]
	y1 = firstEquation[2]
	r1 = firstEquation[3]
	x2 = secondEquation[1]
	y2 = secondEquation[2]
	r2 = secondEquation[3]
	dim temp(3)
	temp[1] =  x1
	temp[2] = -y1
	temp[3] =  r1
	resultY = ((temp[1]*r2) - (x2*temp[3])) / ((x2*temp[2]) + (temp[1]*y2))
	resultX = (r1 - (y1*resultY)) / x1
	print "x = "; resultX
	print "y = "; resultY
end subroutine

call getCrossingPoint(firstEquation, secondEquation)
end
Output:
Igual que la entrada de FreeBASIC.

FreeBASIC[edit]

Dim Shared As Integer firstEquation(1 To 3)  = { 3, 1, -1}
Dim Shared As Integer secondEquation(1 To 3) = { 2,-3,-19}

Sub getCrossingPoint(firstEquation() As Integer, secondEquation() As Integer)
    Dim As Integer x1 = firstEquation(1)
    Dim As Integer y1 = firstEquation(2)
    Dim As Integer r1 = firstEquation(3)
    Dim As Integer x2 = secondEquation(1)
    Dim As Integer y2 = secondEquation(2)
    Dim As Integer r2 = secondEquation(3)
    Dim As Integer temp(3)
    temp(1) =  x1
    temp(2) = -y1
    temp(3) =  r1
    Dim As Integer resultY = ((temp(1)* r2) - (x2 * temp(3))) / ((x2 * temp(2)) + (temp(1)*y2)) 
    Dim As Integer resultX = (r1 - (y1*resultY)) / x1 
    Print "x = "; resultX
	Print "y = "; resultY
End Sub

getCrossingPoint(firstEquation(), secondEquation())
Sleep
Output:
x = -2
y =  5

QBasic[edit]

Works with: QBasic
Works with: QuickBasic
Translation of: FreeBASIC
DIM firstEquation(3)
firstEquation(1) = 3
firstEquation(2) = 1
firstEquation(3) = -1
DIM secondEquation(3)
secondEquation(1) = 2
secondEquation(2) = -3
secondEquation(3) = -19

CALL getCrossingPoint(firstEquation(), secondEquation())
END

SUB getCrossingPoint (firstEquation(), secondEquation())
    x1 = firstEquation(1)
    y1 = firstEquation(2)
    r1 = firstEquation(3)
    x2 = secondEquation(1)
    y2 = secondEquation(2)
    r2 = secondEquation(3)    
    DIM temp(3)
    temp(1) = x1
    temp(2) = -y1
    temp(3) = r1
    resultY = ((temp(1) * r2) - (x2 * temp(3))) / ((x2 * temp(2)) + (temp(1) * y2))
    resultX = (r1 - (y1 * resultY)) / x1
    PRINT "x = "; resultX
    PRINT "y = "; resultY
END SUB
Output:
Igual que la entrada de FreeBASIC.

True BASIC[edit]

Works with: QBasic
Translation of: QBasic
DIM firstequation(3)
LET firstequation(1) = 3
LET firstequation(2) = 1
LET firstequation(3) = -1
DIM secondequation(3)
LET secondequation(1) = 2
LET secondequation(2) = -3
LET secondequation(3) = -19

SUB getcrossingpoint (firstequation(),secondequation())
    LET x1 = firstequation(1)
    LET y1 = firstequation(2)
    LET r1 = firstequation(3)
    LET x2 = secondequation(1)
    LET y2 = secondequation(2)
    LET r2 = secondequation(3)

    DIM temp(3)
    LET temp(1) = x1
    LET temp(2) = -y1
    LET temp(3) = r1

    LET resulty = ((temp(1)*r2)-(x2*temp(3)))/((x2*temp(2))+(temp(1)*y2))
    LET resultx = (r1-(y1*resulty))/x1
    PRINT "x = "; resultx
    PRINT "y = "; resulty
END SUB

CALL getcrossingpoint (firstequation(), secondequation())
END
Output:
Igual que la entrada de FreeBASIC.

Yabasic[edit]

Translation of: FreeBASIC
dim firstEquation(3)
firstEquation(1) = 3
firstEquation(2) = 1
firstEquation(3) = -1
dim secondEquation(3)
secondEquation(1) = 2
secondEquation(2) = -3
secondEquation(3) = -19

sub getCrossingPoint(firstEquation(), secondEquation())
    x1 = firstEquation(1)
    y1 = firstEquation(2)
    r1 = firstEquation(3)
    x2 = secondEquation(1)
    y2 = secondEquation(2)
    r2 = secondEquation(3)
    dim temp(3)
    temp(1) =  x1
    temp(2) = -y1
    temp(3) =  r1
    resultY = ((temp(1)*r2) - (x2*temp(3))) / ((x2*temp(2)) + (temp(1)*y2)) 
    resultX = (r1 - (y1*resultY)) / x1 
    print "x = ", resultX
    print "y = ", resultY
end sub

getCrossingPoint(firstEquation(), secondEquation())
end
Output:
Igual que la entrada de FreeBASIC.

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


Python[edit]

#!/usr/bin/python

firstEquation  = [ 3, 1, -1]
secondEquation = [ 2,-3,-19]

def getCrossingPoint(firstEquation, secondEquation):
    x1 = firstEquation[0]
    y1 = firstEquation[1]
    r1 = firstEquation[2]
    x2 = secondEquation[0]
    y2 = secondEquation[1]
    r2 = secondEquation[2]
    temp = []
    temp.append( x1)
    temp.append(-y1)
    temp.append( r1)
    resultY = ((temp[0]*r2) - (x2*temp[2])) / ((x2*temp[1]) + (temp[0]*y2)) 
    resultX = (r1 - (y1*resultY)) / x1 
    print("x = ", resultX)
    print("y = ", resultY)


if __name__ == "__main__":
    getCrossingPoint(firstEquation, secondEquation)
Output:
x =  -2.0
y =  5.0


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