I would not think so. What makes you think that you would happen to have the solution already in your list of numbers that you're going to make permutations of? If you want to do it numerically, you could use linspace() to make a list of values for x and y, then use meshgrid() to get every (x,y) coordinate in the space, then plug those x,y into your equation, then search for the coordinate that has a value closest to your desired value (10) with min() and ind2sub.
% Make x and y axes.
x=linspace(1,10,100);
y = linspace(1, 20, 100);
% Get x and y over the entire grid.
[xg, yg] = meshgrid(x, y);
% Make the equation for every x and y.
z = 4*xg + 2* yg
% Display it.
image(z);
colormap(gray);
colorbar
% Find where target value lives.
targetValue = 10;
diffMatrix = abs(z - targetValue);
[minDiff, linearIndex] = min(diffMatrix(:))
[row, col] = ind2sub(size(z), linearIndex)
