Solve equations with substitution method
- Task
Let given equations:
3x + y = -1 and 2x - 3y = -19
Solve it with substitution method.
- See related
Julia
<lang 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) </lang>
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
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
<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
XPL0
This shows the vector routines from xpllib.xpl. <lang XPL0>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); ]</lang>
- Output:
x = -2.00000 y = 5.00000