Solve linear Equation System - julia

Question

I have two (mathematical) functions:
y = x
y = -2x + 3
This is solved by y = 1 and x = 1. See picture:
How can I make Julia do this for me?

Answer 1

This is a set of linear equations so first rearrange them in the following way:
-x + y = 0
2x + y = 3
and you see that they are in the form of a linear equation system A*v=b where. A is a matrix:
julia> A = [-1 1; 2 1]
2×2 Array{Int64,2}:
-1 1
2 1
and b is a vector:
julia> b = [0, 3]
2-element Array{Int64,1}:
0
3
Now v contains your unknown variables x and y. You can now solve the system using the left division operator \:
julia> A\b
2-element Array{Float64,1}:
1.0
1.0
If you had a more general system of non-linear equations you should use NLsolve.jl package:
julia> using NLsolve
julia> function f!(F, v)
x = v[1]
y = v[2]
F[1] = -x + y
F[2] = 2*x + y - 3
end
f! (generic function with 1 method)
julia> res = nlsolve(f!, [0.0; 0.0])
Results of Nonlinear Solver Algorithm
* Algorithm: Trust-region with dogleg and autoscaling
* Starting Point: [0.0, 0.0]
* Zero: [1.0000000000003109, 0.9999999999999647]
* Inf-norm of residuals: 0.000000
* Iterations: 2
* Convergence: true
* |x - x'| < 0.0e+00: false
* |f(x)| < 1.0e-08: true
* Function Calls (f): 3
* Jacobian Calls (df/dx): 3
julia> res.zero
2-element Array{Float64,1}:
1.0000000000003109
0.9999999999999647
(note that in f! we define two outputs F[1] and F[2] to be equal to zero - you have to rearrange your equations in this way).
For more details how to use NLsolve.jl see https://github.com/JuliaNLSolvers/NLsolve.jl.

Answer 2

Mr #bogumił-kamiński gave an excellent answer. However, just a friendly reminder, the solution MAY NOT EXIST for some other system of linear equations. In that case, you'll get a SingularException. Consider checking if the solution exists or not. For example,
using LinearAlgebra;
"""
y = x => x - y = 0 => |1 -1| X = |0| => AX = B => X = A⁻¹B
y = -2x + 3 2x + y = 3 |2 1| |3|
"""
function solution()
A::Matrix{Int64} = Matrix{Int64}([1 -1; 2 1]);
br::Matrix{Int64} = Matrix{Int64}([0 3]);
bc = transpose(br);
# bc::Matrix{Int64} = Matrix{Int64}([0 ;3;;]); # I missed a semicolon, that's why I got an error
# println(bc);
if(det(A) == 0) # existence check
println("soln doesnot exist. returning...");
return
end
A⁻¹ = inv(A);
println("solution exists:")
X = A⁻¹ * bc;
println(X);
end
solution();