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.

Task

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

See related

• Reduced row echelon form
• Gaussian elimination

Raku

<lang perl6>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) );</lang>

Output:

(-2 5)

Ring

<lang 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 = []
    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

</lang>

Output:

x = -2
y = 5

Wren

<lang ecmascript>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])")</lang>

Output:

x = -2, y = 5