# Solve equations with substitution method

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

See related

## AWK

```# 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

### BASIC256

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

```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

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

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

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

```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

```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[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

```#!/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

```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[1] y1 = firstEquation[2] r1 = firstEquation[3] x2 = secondEquation[1] y2 = secondEquation[2] r2 = secondEquation[3]
temp = []
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

```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

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
```