-- f just generates all possible lists of length k where each element is at least as great as the next element.
-- I wanted to do this efficient-like, which is why this may be a little confusing.
f min n 0 = [[]]
f min n k = foldr (++) [] [[a:as | as <- (f a n (k-1))] | a <- [min..n]]
-- If I had done it the easy way instead of the efficient way using f
-- then then I would have just let x = [[a,b,c,d] | a<-[1..20],b<-[1..20],c<-[1..20],a*b+c*d==a+b+c+d+3 && a >= b && b >= c && c >= d]
-- and not defined f or y
y = f 1 20 4
x = [[a,b,c,d] | [a,b,c,d] <- y, a*b+c*d==a+b+c+d+3]The first step is to isolate the variable that you want to solve for on one side of the equation. Then, use inverse operations to solve for that variable. Repeat this process for each variable until all four have been solved for.
Start by solving for the most isolated variable. This means the variable that has the least number of terms or factors attached to it. If two variables are equally isolated, choose the one that has the smallest coefficient (number in front of the variable).
Yes, it is possible to solve for all four variables at the same time. However, this may require more advanced algebraic techniques such as substitution or elimination. It is generally easier to solve for one variable at a time.
Some common mistakes include not distributing correctly, forgetting to apply the distributive property, and making a sign error when combining like terms. It is also important to check your answer by plugging it back into the original equation to ensure it satisfies all parts of the equation.
One helpful tip is to label each variable with a different letter to keep track of which variable you are solving for. Another tip is to simplify the equation as much as possible before starting to solve for the variables. It may also be helpful to practice with simpler equations before moving on to more complex ones.