Creating an algorithm for a Simultaneous Equation
6 次查看(过去 30 天)
显示 更早的评论
3000x + 1000y = 35000 (1)
240x + 5y = 1665 (2)
Please can you help me create an algorithm that will keep increasing the value of 3000 by 20% (i.e 3000 + 20%*3000) and at the same time giving the corresponding results of "x" and "y".
Thank you in anticipation.
0 个评论
采纳的回答
David Hill
2021-3-31
k=3000*(.2).^(0:10);%however long you want
for j=1:length(k)
A=[k(j) 1000;240 5];
b=[35000;1665];
out(j,:)=A\b;
end
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Numerical Integration and Differential Equations 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!